Consider the following L2-ODE-CC:
y
″
−
10
y
′
+
25
y
=
7
e
5
x
−
2
x
2
{\displaystyle \displaystyle y^{''}-10y^{'}+25y=7e^{5x}-2x^{2}}
General solution of our ODE:
y
″
−
10
y
′
+
25
y
=
0
{\displaystyle \displaystyle y^{''}-10y^{'}+25y=0}
a
2
−
4
b
=
10
2
−
4
(
25
)
=
0
⇒
{\displaystyle \displaystyle a^{2}-4b=10^{2}-4(25)=0\Rightarrow }
λ
=
λ
1
=
λ
2
=
−
a
2
=
−
−
10
2
=
5
{\displaystyle \displaystyle \lambda =\lambda _{1}=\lambda _{2}=-{\frac {a}{2}}=-{\frac {-10}{2}}=5}
y
g
=
(
c
1
+
c
2
x
)
e
5
x
{\displaystyle \displaystyle y_{g}=(c_{1}+c_{2}x)e^{5x}}
r
(
x
)
=
k
e
γ
x
−
k
x
n
{\displaystyle \displaystyle r(x)=ke^{\gamma x}-kx^{n}}
y
p
1
=
C
x
2
e
γ
x
,
y
p
2
=
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle \displaystyle y_{p1}=Cx^{2}e^{\gamma x},y_{p2}=K_{2}x^{2}+K_{1}x+K_{0}}
Have to multiply
y
p
1
{\displaystyle \displaystyle y_{p1}}
x
2
{\displaystyle \displaystyle x^{2}}
e
5
x
{\displaystyle \displaystyle e^{5x}}
Taking derivatives of
y
p
1
{\displaystyle \displaystyle y_{p1}}
y
p
1
=
C
x
2
e
5
x
{\displaystyle \displaystyle y_{p1}=Cx^{2}e^{5x}}
y
p
1
′
=
2
C
x
e
5
x
+
5
C
x
2
e
5
x
{\displaystyle \displaystyle y_{p1}^{'}=2Cxe^{5x}+5Cx^{2}e^{5x}}
y
p
1
″
=
2
C
e
5
x
+
10
C
x
e
5
x
+
10
C
x
e
5
x
+
25
C
x
2
e
5
x
{\displaystyle \displaystyle y_{p1}^{''}=2Ce^{5x}+10Cxe^{5x}+10Cxe^{5x}+25Cx^{2}e^{5x}}
C
[
2
+
10
x
+
10
x
+
25
x
2
−
20
x
−
50
x
2
+
25
x
2
]
e
5
x
=
7
e
5
x
{\displaystyle \displaystyle C[2+10x+10x+25x^{2}-20x-50x^{2}+25x^{2}]e^{5x}=7e^{5x}}
2
C
=
7
⇒
C
=
7
2
{\displaystyle \displaystyle 2C=7\Rightarrow C={\frac {7}{2}}}
y
p
1
=
7
2
x
2
e
5
x
{\displaystyle \displaystyle y_{p1}={\frac {7}{2}}x^{2}e^{5x}}
y
p
2
{\displaystyle \displaystyle y_{p2}}
y
p
2
=
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle \displaystyle y_{p2}=K_{2}x^{2}+K_{1}x+K_{0}}
y
p
2
′
=
2
K
2
x
+
K
1
{\displaystyle \displaystyle y_{p2}^{'}=2K_{2}x+K_{1}}
y
p
2
″
=
2
K
2
{\displaystyle \displaystyle y_{p2}^{''}=2K_{2}}
25
K
2
=
−
2
⇒
K
2
=
−
2
25
{\displaystyle \displaystyle 25K_{2}=-2\Rightarrow K_{2}=-{\frac {2}{25}}}
−
20
K
2
+
25
K
1
=
0
⇒
K
1
=
−
8
125
{\displaystyle \displaystyle -20K_{2}+25K_{1}=0\Rightarrow K_{1}=-{\frac {8}{125}}}
2
K
2
−
10
K
1
+
25
K
0
=
0
⇒
K
0
=
−
12
625
{\displaystyle \displaystyle 2K_{2}-10K_{1}+25K_{0}=0\Rightarrow K_{0}=-{\frac {12}{625}}}
y
p
2
=
−
2
25
x
2
−
8
125
x
−
12
625
{\displaystyle \displaystyle y_{p2}=-{\frac {2}{25}}x^{2}-{\frac {8}{125}}x-{\frac {12}{625}}}
y
=
y
g
+
y
p
1
+
y
p
2
{\displaystyle \displaystyle y=y_{g}+y_{p1}+y_{p2}}
y
=
(
c
1
+
c
2
x
)
e
5
x
+
7
2
x
2
e
5
x
−
2
25
x
2
−
8
125
x
−
12
625
{\displaystyle \displaystyle y=(c1+c2x)e^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}-{\frac {8}{125}}x-{\frac {12}{625}}}
y
(
0
)
=
4
⇒
4
=
c
1
−
12
625
⇒
c
1
=
2512
625
{\displaystyle \displaystyle y(0)=4\Rightarrow 4=c_{1}-{\frac {12}{625}}\Rightarrow c_{1}={\frac {2512}{625}}}
y
′
(
0
)
=
−
5
{\displaystyle \displaystyle y^{'}(0)=-5}
y
′
=
e
5
x
[
5
c
1
+
c
2
+
5
c
2
x
+
7
x
+
35
2
x
2
]
−
4
25
x
−
8
125
{\displaystyle \displaystyle y^{'}=e^{5x}[5c_{1}+c_{2}+5c_{2}x+7x+{\frac {35}{2}}x^{2}]-{\frac {4}{25}}x-{\frac {8}{125}}}
−
5
=
5
c
1
+
c
2
−
8
125
⇒
c
2
=
−
3129
125
{\displaystyle \displaystyle -5=5c_{1}+c_{2}-{\frac {8}{125}}\Rightarrow c_{2}=-{\frac {3129}{125}}}
y
=
(
2512
625
−
3129
125
x
)
e
5
x
+
7
2
x
2
e
5
x
−
2
25
x
2
−
8
125
x
−
12
625
{\displaystyle \displaystyle y=({\frac {2512}{625}}-{\frac {3129}{125}}x)e^{5x}+{\frac {7}{2}}x^{2}e^{5x}-{\frac {2}{25}}x^{2}-{\frac {8}{125}}x-{\frac {12}{625}}}
x=0:0.001:10;
y=((2512/625)-(3129/125).*x).*exp(5*x)+(7/2).*x.^2.*exp(5*x)-(2/25).*x.^2-(8/125).*x-(12/625);
plot(x,y),xlabel('x'),ylabel('y(x)')
y
=
4
e
5
x
−
25
x
e
5
x
+
7
2
x
2
e
5
x
{\displaystyle \displaystyle y=4e^{5x}-25xe^{5x}+{\frac {7}{2}}x^{2}e^{5x}}
Matlab code:
x=0:0.001:10;
y=4.*exp(5*x)-25.*x.*exp(5*x)+(7/2).*x.^2.*exp(5*x);
plot(x,y),xlabel('x'),ylabel('y(x)')
This problem was solved and uploaded by: Joshua House
This problem was proofread by: David Herrick
Find the homogeneous L2-ODE-CC having the following roots:
λ
1
=
λ
,
λ
2
=
λ
+
ε
{\displaystyle \displaystyle \lambda _{1}=\lambda ,\lambda _{2}=\lambda +\varepsilon }
Show that the following is a homogeneous solution:
e
(
λ
+
ε
)
x
−
e
λ
x
ε
{\displaystyle \displaystyle {\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }}}
Find the limit of the homogeneous solution as
ε
→
0
{\displaystyle \displaystyle \varepsilon \to 0}
Take the derivative of
e
λ
x
{\displaystyle \displaystyle e^{\lambda x}}
λ
{\displaystyle \displaystyle \lambda }
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters
Compute (2) using
λ
=
5
{\displaystyle \displaystyle \lambda =5}
ε
=
0.001
{\displaystyle \displaystyle \varepsilon =0.001}
Letting
λ
1
=
x
,
λ
2
=
x
+
ϵ
{\displaystyle \displaystyle \lambda _{1}=x,\lambda _{2}=x+\epsilon }
(
λ
−
x
)
(
λ
−
[
x
+
ε
]
)
=
0
{\displaystyle \displaystyle (\lambda -x)(\lambda -[x+\varepsilon ])=0}
λ
2
−
λ
x
−
λ
ε
−
λ
x
+
x
2
+
x
ε
=
0
{\displaystyle \displaystyle \lambda ^{2}-\lambda x-\lambda \varepsilon -\lambda x+x^{2}+x\varepsilon =0}
λ
2
−
λ
[
2
x
+
ε
]
+
x
[
x
+
ε
]
=
0
{\displaystyle \displaystyle \lambda ^{2}-\lambda [2x+\varepsilon ]+x[x+\varepsilon ]=0}
Replacing the x's with lambda's:
y
″
−
y
′
[
2
λ
+
ε
]
+
y
λ
[
λ
+
ε
]
=
0
{\displaystyle \displaystyle y^{''}-y^{'}[2\lambda +\varepsilon ]+y\lambda [\lambda +\varepsilon ]=0}
y
=
e
(
λ
+
ε
)
x
−
e
λ
x
ε
,
y
′
=
(
λ
+
ε
)
e
(
λ
+
ε
)
x
−
λ
e
λ
x
ε
,
y
″
=
(
λ
+
ε
)
2
e
(
λ
+
ε
)
x
−
λ
2
e
λ
x
ε
{\displaystyle \displaystyle y={\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }},y'={\frac {(\lambda +\varepsilon )e^{(\lambda +\varepsilon )x}-\lambda e^{\lambda x}}{\varepsilon }},y''={\frac {(\lambda +\varepsilon )^{2}e^{(\lambda +\varepsilon )x}-\lambda ^{2}e^{\lambda x}}{\varepsilon }}}
y
″
−
y
′
[
2
λ
+
ε
]
+
y
λ
[
λ
+
ε
]
=
0
{\displaystyle \displaystyle y^{''}-y^{'}[2\lambda +\varepsilon ]+y\lambda [\lambda +\varepsilon ]=0}
e
(
λ
+
ε
)
x
[
λ
2
+
2
λ
ε
+
ε
2
−
2
λ
2
−
2
λ
ε
−
λ
ε
−
ε
2
+
λ
2
+
ε
λ
]
+
e
λ
x
[
−
λ
2
+
2
λ
2
+
λ
ε
−
λ
2
−
λ
ε
]
=
0
{\displaystyle \displaystyle e^{(\lambda +\varepsilon )x}[\lambda ^{2}+2\lambda \varepsilon +\varepsilon ^{2}-2\lambda ^{2}-2\lambda \varepsilon -\lambda \varepsilon -\varepsilon ^{2}+\lambda ^{2}+\varepsilon \lambda ]+e^{\lambda x}[-\lambda ^{2}+2\lambda ^{2}+\lambda \varepsilon -\lambda ^{2}-\lambda \varepsilon ]=0}
Much of the trivial algebra was left out in order to greatly reduce coding time. These intermediate steps, which were done on paper, can be presented if necessary.
lim
ε
→
0
e
(
λ
+
ε
)
x
−
e
λ
x
ε
{\displaystyle \displaystyle \lim _{\varepsilon \rightarrow 0}{\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }}}
If:
lim
ε
→
0
f
′
(
ε
)
g
′
(
ε
)
{\displaystyle \displaystyle \lim _{\varepsilon \rightarrow 0}{\frac {f'(\varepsilon )}{g'(\varepsilon )}}}
lim
ε
→
0
f
(
ε
)
=
lim
ε
→
0
g
(
ε
)
=
0
{\displaystyle \displaystyle \lim _{\varepsilon \rightarrow 0}f(\varepsilon )=\lim _{\varepsilon \rightarrow 0}g(\varepsilon )=0}
lim
ε
→
0
f
(
ε
)
g
(
ε
)
=
lim
ε
→
0
f
′
(
ε
)
g
′
(
ε
)
{\displaystyle \displaystyle \lim _{\varepsilon \rightarrow 0}{\frac {f(\varepsilon )}{g(\varepsilon )}}=\lim _{\varepsilon \rightarrow 0}{\frac {f'(\varepsilon )}{g'(\varepsilon )}}}
lim
ε
→
0
x
e
(
λ
+
ε
)
x
1
=
x
e
λ
x
{\displaystyle \displaystyle \lim _{\varepsilon \rightarrow 0}{\frac {xe^{(\lambda +\varepsilon )x}}{1}}=xe^{\lambda x}}
d
[
e
λ
x
]
d
λ
=
x
e
λ
x
{\displaystyle \displaystyle {\frac {d[e^{\lambda x}]}{d\lambda }}=xe^{\lambda x}}
The results from (3) and (4) show that
d
[
e
λ
x
]
d
λ
=
lim
e
→
0
e
(
λ
+
ε
)
x
−
e
λ
x
ε
=
x
e
λ
x
{\displaystyle \displaystyle {\frac {d[e^{\lambda x}]}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }}=xe^{\lambda x}}
From
e
(
λ
+
ε
)
x
−
e
λ
x
ε
{\displaystyle \displaystyle {\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }}}
λ
=
5
,
ε
=
0.001
{\displaystyle \displaystyle \lambda =5,\varepsilon =0.001}
e
5.001
x
−
e
5
x
0.001
{\displaystyle \displaystyle {\frac {e^{5.001x}-e^{5x}}{0.001}}}
e
5.001
x
[
5
2
+
2
(
5
)
(
0.001
)
+
0.001
2
−
2
(
5
2
)
−
2
(
5
)
(
0.001
)
−
5
(
0.001
)
−
(
0.001
2
)
+
5
2
+
0.001
(
5
)
]
+
e
5
x
[
−
5
2
+
2
(
5
2
)
+
5
(
0.001
)
−
5
2
−
5
(
0.001
)
]
=
0
{\displaystyle \displaystyle e^{5.001x}[5^{2}+2(5)(0.001)+0.001^{2}-2(5^{2})-2(5)(0.001)-5(0.001)-(0.001^{2})+5^{2}+0.001(5)]+e^{5x}[-5^{2}+2(5^{2})+5(0.001)-5^{2}-5(0.001)]=0}
This problem was solved and uploaded by: Joshua House
This problem was proofread by: Michael Wallace
Find the complete solution for equation 5 from pg. 7.7 in the notes:
y
″
−
2
y
′
+
3
y
=
4
x
2
{\displaystyle \displaystyle \ y''-2y'+3y=4x^{2}\ }
with initial conditions given as:
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \displaystyle \ y(0)=1,y'(0)=0\ }
Plot the solution.
First, find the homogenous solution to this differential equation, i.e. where r(x) = 0.
y
h
″
−
3
y
h
′
+
2
y
h
=
0
{\displaystyle \displaystyle \ y''_{h}-3y'_{h}+2y_{h}=0\ }
(
λ
−
1
)
(
λ
−
2
)
=
0
{\displaystyle \displaystyle \ (\lambda -1)(\lambda -2)=0\ }
λ
=
1
,
2
{\displaystyle \displaystyle \ \lambda =1,2\ }
Thus, the homogenous solution is of the form
y
h
=
C
1
e
λ
1
x
+
C
2
e
λ
2
x
{\displaystyle \displaystyle \ y_{h}=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}x}\ }
Plugging in the answers for lambda yields:
y
h
=
C
1
e
x
+
C
2
e
2
x
{\displaystyle \displaystyle \ y_{h}=C_{1}e^{x}+C_{2}e^{2x}\ }
Now to solve for the particular solution, we must look at the excitation:
r
(
x
)
=
4
x
2
{\displaystyle \displaystyle \ r(x)=4x^{2}\ }
A particular solution to an excitation of the form
C
x
n
{\displaystyle \displaystyle \ Cx^{n}\ }
∑
j
=
0
n
c
j
x
j
{\displaystyle \displaystyle \ \sum _{j=0}^{n}c_{j}x^{j}\ }
Thus,
y
p
=
c
0
+
c
1
x
+
c
2
x
2
{\displaystyle \displaystyle \ y_{p}=c_{0}+c_{1}x+c_{2}x^{2}\ }
To solve for the 3 unknown constants, plug the particular solution into the differential equation:
y
p
″
−
3
y
p
′
+
2
y
p
=
4
x
2
{\displaystyle \displaystyle \ y_{p}''-3y_{p}'+2y_{p}=4x^{2}\ }
(
c
0
+
c
1
x
+
c
2
x
2
)
″
−
3
(
c
0
+
c
1
x
+
c
2
x
2
)
′
+
2
(
c
0
+
c
1
x
+
c
2
x
2
)
=
4
x
2
{\displaystyle \displaystyle \ (c_{0}+c_{1}x+c_{2}x^{2})''-3(c_{0}+c_{1}x+c_{2}x^{2})'+2(c_{0}+c_{1}x+c_{2}x^{2})=4x^{2}\ }
y
p
′
=
2
c
2
x
+
c
1
{\displaystyle \displaystyle \ y_{p}'=2c_{2}x+c_{1}\ }
y
p
″
=
2
c
2
{\displaystyle \displaystyle \ y_{p}''=2c_{2}\ }
Substituting these relations into the equation we get:
2
c
2
−
3
(
2
c
2
x
+
c
1
)
+
2
(
c
2
x
2
+
c
1
x
+
c
0
)
=
4
x
2
{\displaystyle \displaystyle \ 2c_{2}-3(2c_{2}x+c_{1})+2(c_{2}x^{2}+c_{1}x+c_{0})=4x^{2}\ }
Grouping like terms we get:
2
c
2
x
2
+
(
2
c
1
−
6
c
2
)
x
+
(
2
c
2
−
3
c
1
+
2
c
0
)
=
4
x
2
{\displaystyle \displaystyle \ 2c_{2}x^{2}+(2c_{1}-6c_{2})x+(2c_{2}-3c_{1}+2c_{0})=4x^{2}\ }
Matching up the coefficients for the
x
2
{\displaystyle \displaystyle \ x^{2}\ }
2
c
2
=
4
{\displaystyle \displaystyle \ 2c_{2}=4\ }
Therefore
c
2
=
2
{\displaystyle \displaystyle \ c_{2}=2\ }
Matching up the coefficients for the
x
{\displaystyle \displaystyle \ x\ }
2
c
1
−
6
c
2
=
0
{\displaystyle \displaystyle \ 2c_{1}-6c_{2}=0\ }
Therefore
c
1
=
6
c
2
/
2
=
6
(
2
)
/
2
=
6
{\displaystyle \displaystyle \ c_{1}=6c_{2}/2=6(2)/2=6\ }
Finally, matching up the coefficients for the constant terms on the left and right sides we get:
2
c
2
−
3
c
1
+
2
c
0
=
0
{\displaystyle \displaystyle \ 2c_{2}-3c_{1}+2c_{0}=0\ }
Therefore
c
0
=
(
3
c
1
−
2
c
2
)
/
2
=
(
3
(
6
)
−
2
(
2
)
)
/
2
=
(
18
−
4
)
/
2
=
7
{\displaystyle \displaystyle \ c_{0}=(3c_{1}-2c_{2})/2=(3(6)-2(2))/2=(18-4)/2=7\ }
The particular solution is then:
y
p
=
2
x
2
+
6
x
+
7
{\displaystyle \displaystyle \ y_{p}=2x^{2}+6x+7\ }
The general solution is the sum of the homogenous solution and the particular solution, therefore:
y
(
x
)
=
y
h
+
y
p
=
C
1
e
x
+
C
2
e
2
x
+
2
x
2
+
6
x
+
7
{\displaystyle \displaystyle \ y(x)=y_{h}+y_{p}=C_{1}e^{x}+C_{2}e^{2x}+2x^{2}+6x+7\ }
To solve for the 2 remaining unknown constants, we use our initial conditions from the problem statement.
y
(
0
)
=
1
=
C
1
+
C
2
+
7
{\displaystyle \displaystyle \ y(0)=1=C_{1}+C_{2}+7\ }
To use the second initial condition, we must take a derivative of our general solution.
y
′
(
x
)
=
C
1
e
x
+
2
C
2
e
2
x
+
4
x
+
6
{\displaystyle \displaystyle \ y'(x)=C_{1}e^{x}+2C_{2}e^{2x}+4x+6\ }
y
′
(
0
)
=
0
=
C
1
+
2
C
2
+
6
{\displaystyle \displaystyle \ y'(0)=0=C_{1}+2C_{2}+6\ }
Rewriting the equations we get:
C
1
+
2
C
2
+
6
=
0
{\displaystyle \displaystyle \ C_{1}+2C_{2}+6=0\ }
C
1
+
C
2
+
6
=
0
{\displaystyle \displaystyle \ C_{1}+C_{2}+6=0\ }
Subtracting these equations we get:
C
2
=
0
{\displaystyle \displaystyle \ C_{2}=0\ }
Therefore:
C
1
=
−
6
−
C
2
=
−
6
{\displaystyle \displaystyle \ C_{1}=-6-C_{2}=-6\ }
Thus the complete solution is:
y
(
x
)
=
2
x
2
+
6
x
+
7
−
6
e
x
{\displaystyle \displaystyle \ y(x)=2x^{2}+6x+7-6e^{x}\ }
MATLAB code;
x = 0:0.001:10;
y = 2*x.^2 + 6*x + 7 - 6*exp(x);
plot(x,y)
xlabel( 'x')
ylabel( 'y(x)')
This problem was solved and uploaded by: David Herrick
This problem was proofread by: Michael Wallace
Use the Basic Rule (1) and the Sum Rule (3) on p.7-2 of the notes to show that the appropriate particular solution for:
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle \ y''-3y'+2y=4x^{2}-6x^{5}\ }
is of the form:
y
p
(
x
)
=
∑
j
=
0
5
c
j
x
j
{\displaystyle \ y_{p}(x)=\sum _{j=0}^{5}c_{j}x^{j}\ }
with n = 5, i.e. (1) on p.7-12.
From the Basic Rule:
r
1
(
x
)
=
4
x
2
{\displaystyle \ r_{1}(x)=4x^{2}\ }
r
2
(
x
)
=
6
x
5
{\displaystyle \ r_{2}(x)=6x^{5}\ }
Therefore, from Table 2.1:
y
p
1
(
x
)
=
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle y_{p_{1}}(x)=K_{2}x^{2}+K_{1}x+K_{0}}
y
p
2
(
x
)
=
K
5
x
5
+
K
4
x
4
+
K
3
x
3
+
K
2
x
2
+
K
1
x
+
K
0
{\displaystyle y_{p_{2}}(x)=K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0}}
For
y
p
1
(
x
)
{\displaystyle y_{p_{1}}(x)}
y
p
1
′
(
x
)
=
2
K
2
x
+
K
1
{\displaystyle \ y_{p_{1}}'(x)=2K_{2}x+K_{1}\ }
y
p
1
″
(
x
)
=
2
K
2
{\displaystyle \ y_{p_{1}}''(x)=2K_{2}\ }
Substituting
y
p
1
(
x
)
{\displaystyle y_{p_{1}}(x)}
(
2
K
2
)
−
3
(
2
K
2
x
+
K
1
)
+
2
(
K
2
x
2
+
K
1
x
+
K
0
)
=
4
x
2
−
6
x
5
{\displaystyle \ (2K_{2})-3(2K_{2}x+K_{1})+2(K_{2}x^{2}+K_{1}x+K_{0})=4x^{2}-6x^{5}\ }
=
x
2
(
2
K
2
)
+
x
(
2
K
1
+
3
K
2
)
+
(
2
K
2
+
3
K
1
+
2
K
0
)
=
4
x
2
−
6
x
5
{\displaystyle \ =x^{2}(2K_{2})+x(2K_{1}+3K_{2})+(2K_{2}+3K_{1}+2K_{0})=4x^{2}-6x^{5}\ }
Comparing the
x
2
{\displaystyle x^{2}}
x
{\displaystyle x}
x
0
{\displaystyle x^{0}}
x
2
{\displaystyle x^{2}}
2
K
2
=
4
{\displaystyle \ 2K_{2}=4\ }
K
2
=
2
{\displaystyle K_{2}=2}
x
{\displaystyle x}
2
K
1
+
3
K
2
=
0
{\displaystyle \ 2K_{1}+3K_{2}=0\ }
K
1
=
−
3
{\displaystyle K_{1}=-3}
x
0
{\displaystyle x^{0}}
2
K
2
+
3
K
1
+
2
K
0
=
0
{\displaystyle \ 2K_{2}+3K_{1}+2K_{0}=0\ }
K
0
=
2.5
{\displaystyle K_{0}=2.5}
y
p
1
(
x
)
=
2
x
2
−
3
x
+
2.5
{\displaystyle y_{p_{1}}(x)=2x^{2}-3x+2.5}
y
p
2
(
x
)
{\displaystyle y_{p_{2}}(x)}
y
p
2
′
(
x
)
=
5
K
5
x
4
+
4
K
4
x
3
+
3
K
3
x
2
+
2
K
2
x
+
K
1
{\displaystyle \ y_{p_{2}}'(x)=5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1}\ }
y
p
2
″
(
x
)
=
20
K
5
x
3
+
12
K
4
x
2
+
6
K
3
x
+
2
K
2
{\displaystyle \ y_{p_{2}}''(x)=20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2}\ }
Substituting
y
p
1
(
x
)
{\displaystyle y_{p_{1}}(x)}
(
20
K
5
x
3
+
12
K
4
x
2
+
6
K
3
x
+
2
K
2
)
−
3
(
5
K
5
x
4
+
4
K
4
x
3
+
3
K
3
x
2
+
2
K
2
x
+
K
1
)
+
2
(
K
5
x
5
+
K
4
x
4
+
K
3
x
3
+
K
2
x
2
+
K
1
x
+
K
0
)
=
4
x
2
−
6
x
5
{\displaystyle \ (20K_{5}x^{3}+12K_{4}x^{2}+6K_{3}x+2K_{2})-3(5K_{5}x^{4}+4K_{4}x^{3}+3K_{3}x^{2}+2K_{2}x+K_{1})+2(K_{5}x^{5}+K_{4}x^{4}+K_{3}x^{3}+K_{2}x^{2}+K_{1}x+K_{0})=4x^{2}-6x^{5}\ }
Simplifying yields:
x
5
(
K
5
)
+
x
4
(
K
4
−
15
K
5
)
+
x
3
(
K
3
−
12
K
4
+
40
K
5
)
+
x
2
(
K
2
−
9
K
3
+
24
K
4
)
+
x
(
K
1
−
6
K
2
+
12
K
3
)
+
(
K
0
−
3
K
1
+
4
K
2
)
=
4
x
2
−
6
x
5
{\displaystyle \ x^{5}(K_{5})+x^{4}(K_{4}-15K_{5})+x^{3}(K_{3}-12K_{4}+40K_{5})+x^{2}(K_{2}-9K_{3}+24K_{4})+x(K_{1}-6K_{2}+12K_{3})+(K_{0}-3K_{1}+4K_{2})=4x^{2}-6x^{5}\ }
Comparing the
x
5
{\displaystyle x^{5}}
x
4
{\displaystyle x^{4}}
x
3
{\displaystyle x^{3}}
x
2
{\displaystyle x^{2}}
x
{\displaystyle x}
x
0
{\displaystyle x^{0}}
x
5
{\displaystyle x^{5}}
K
5
=
−
6
{\displaystyle \ K_{5}=-6\ }
K
5
=
−
6
{\displaystyle K_{5}=-6}
x
4
{\displaystyle x^{4}}
K
4
−
15
K
5
=
0
{\displaystyle \ K_{4}-15K_{5}=0\ }
K
4
=
−
90
{\displaystyle K_{4}=-90}
x
3
{\displaystyle x^{3}}
K
3
−
12
K
4
+
40
K
5
=
0
{\displaystyle \ K_{3}-12K_{4}+40K_{5}=0\ }
K
3
=
840
{\displaystyle K_{3}=840}
x
2
{\displaystyle x^{2}}
K
2
−
9
K
3
+
24
K
4
=
0
{\displaystyle \ K_{2}-9K_{3}+24K_{4}=0\ }
K
2
=
−
9724
{\displaystyle K_{2}=-9724}
x
{\displaystyle x}
K
1
−
6
K
2
+
12
K
3
=
0
{\displaystyle \ K_{1}-6K_{2}+12K_{3}=0\ }
K
1
=
59425
{\displaystyle K_{1}=59425}
x
0
{\displaystyle x^{0}}
K
0
−
3
K
1
+
4
K
2
=
0
{\displaystyle \ K_{0}-3K_{1}+4K_{2}=0\ }
K
0
=
−
217171
{\displaystyle K_{0}=-217171}
y
p
2
(
x
)
=
−
6
x
5
−
90
x
4
+
840
x
3
−
9724
x
2
+
59425
x
−
217171
{\displaystyle y_{p_{2}}(x)=-6x^{5}-90x^{4}+840x^{3}-9724x^{2}+59425x-217171}
y
p
(
x
)
=
y
p
1
(
x
)
+
y
p
2
(
x
)
{\displaystyle \ y_{p}(x)=y_{p_{1}}(x)+y_{p_{2}}(x)\ }
y
p
(
x
)
=
(
2
x
2
−
3
x
+
2.5
)
+
(
−
6
x
5
−
90
x
4
+
840
x
3
−
9724
x
2
+
59425
x
−
217171
)
{\displaystyle \ y_{p}(x)=(2x^{2}-3x+2.5)+(-6x^{5}-90x^{4}+840x^{3}-9724x^{2}+59425x-217171)\ }
Simplifying gives:
y
p
(
x
)
=
−
6
x
5
−
90
x
4
+
840
x
3
−
9722
x
2
+
59422
x
−
217168.5
{\displaystyle \ y_{p}(x)=-6x^{5}-90x^{4}+840x^{3}-9722x^{2}+59422x-217168.5\ }
Choose:
K
5
=
−
6
{\displaystyle \ K_{5}=-6\ }
K
4
=
−
90
{\displaystyle \ K_{4}=-90\ }
K
3
=
840
{\displaystyle \ K_{3}=840\ }
K
2
=
−
9722
{\displaystyle \ K_{2}=-9722\ }
K
1
=
59422
{\displaystyle \ K_{1}=59422\ }
K
0
=
−
217168.5
{\displaystyle \ K_{0}=-217168.5\ }
Then the equation becomes of the form:
y
p
(
x
)
=
∑
j
=
0
n
(
c
j
x
j
)
{\displaystyle \ y_{p}(x)=\sum _{j=0}^{n}(c_{j}x^{j})\ }
This problem was solved and uploaded by John North .
This problem was proofread by Michael Wallace
Complete the solution for
y
″
−
3
y
′
+
2
y
=
4
x
2
−
6
x
5
{\displaystyle \ y''-3y'+2y=4x^{2}-6x^{5}\ }
1) Obtain equations (2) - (4) and (6) p. 7-14
(2) Coefficients of
x
{\displaystyle \ x\ }
(3) Coefficients of
x
2
{\displaystyle \ x^{2}\ }
(4) Coefficients of
x
3
{\displaystyle \ x^{3}\ }
(6) Coefficients of
x
5
{\displaystyle \ x^{5}\ }
Verify all equations by long hand expansion of the series in (4) p. 7-12, instead of using the series in (2) p. 7-13.
∑
j
=
2
5
c
j
⋅
j
⋅
(
j
−
1
)
⋅
x
j
−
2
−
3
∑
j
=
1
5
c
j
⋅
j
⋅
x
j
−
1
+
2
∑
j
=
0
5
c
j
x
j
=
4
x
2
−
6
x
5
{\displaystyle \ \sum _{j=2}^{5}c_{j}\cdot j\cdot (j-1)\cdot x^{j-2}-3\sum _{j=1}^{5}c_{j}\cdot j\cdot x^{j-1}+2\sum _{j=0}^{5}c_{j}x^{j}=4x^{2}-6x^{5}\ }
Put the system of equations for
{
c
0
,
.
.
.
,
c
5
}
{\displaystyle \ \left\{c_{0},...,c_{5}\right\}\ }
Solve for the coefficients
{
c
0
,
.
.
.
,
c
5
}
{\displaystyle \ \left\{c_{0},...,c_{5}\right\}\ }
Consider the initial conditions:
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \ y(0)=1,y'(0)=0\ }
Find the solution y(x) and plot it.
For all the coefficient equations, we will use the series from (2) p. 7-13
∑
j
=
0
3
[
c
j
+
2
(
j
+
2
)
(
j
+
1
)
−
3
c
j
+
1
(
j
+
1
)
+
2
c
j
]
x
j
−
3
c
5
(
5
)
x
4
+
2
[
c
4
x
4
+
c
5
x
5
]
=
4
x
2
−
6
x
5
{\displaystyle \ \sum _{j=0}^{3}\left[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}\right]x^{j}-3c_{5}(5)x^{4}+2\left[c_{4}x^{4}+c_{5}x^{5}\right]=4x^{2}-6x^{5}\ }
Solving for the coefficients of x, we use j = 1.
(
c
3
(
3
)
(
2
)
−
3
c
2
(
2
)
+
2
c
1
)
x
{\displaystyle \ (c_{3}(3)(2)-3c_{2}(2)+2c_{1})x\ }
These are the coefficients for x on the left hand side of the equation, and the coefficients for x on the right hand side = 0.
Coefficients of x:
6
c
3
−
6
c
2
+
2
c
1
=
0
{\displaystyle \ 6c_{3}-6c_{2}+2c_{1}=0\ }
Solving for the coefficients of
x
2
{\displaystyle \ x^{2}\ }
(
c
4
(
4
)
(
3
)
−
3
c
3
(
3
)
+
2
c
2
)
x
2
{\displaystyle \ (c_{4}(4)(3)-3c_{3}(3)+2c_{2})x^{2}\ }
These are the coefficients for
x
2
{\displaystyle \ x^{2}\ }
x
2
{\displaystyle \ x^{2}\ }
Coefficients of
x
2
{\displaystyle \ x^{2}\ }
12
c
4
−
9
c
3
+
2
c
2
=
4
{\displaystyle \ 12c_{4}-9c_{3}+2c_{2}=4\ }
Solving for the coefficients of
x
3
{\displaystyle \ x^{3}\ }
(
c
5
(
5
)
(
4
)
−
3
c
4
(
4
)
+
2
c
3
)
x
3
{\displaystyle \ (c_{5}(5)(4)-3c_{4}(4)+2c_{3})x^{3}\ }
These are the coefficients for
x
3
{\displaystyle \ x^{3}\ }
x
3
{\displaystyle \ x^{3}\ }
Coefficients of
x
3
{\displaystyle \ x^{3}\ }
20
c
5
−
12
c
4
+
2
c
3
=
0
{\displaystyle \ 20c_{5}-12c_{4}+2c_{3}=0\ }
Solving for the coefficients of
x
5
{\displaystyle \ x^{5}\ }
2
c
5
x
5
{\displaystyle \ 2c_{5}x^{5}\ }
This is the only coefficient of
x
5
{\displaystyle \ x^{5}\ }
x
5
{\displaystyle \ x^{5}\ }
Coefficients of
x
5
{\displaystyle \ x^{5}\ }
2
c
5
=
−
6
{\displaystyle \ 2c_{5}=-6\ }
Expanding the 3 series using j = 0 to 5, we get:
2
c
0
+
2
c
1
x
−
3
c
1
+
2
c
2
x
2
−
3
c
2
(
2
)
x
+
2
c
2
+
2
c
3
x
3
−
3
(
3
)
c
3
x
2
+
3
(
2
)
c
3
x
+
2
c
4
x
4
−
3
(
4
)
c
4
x
3
+
4
(
3
)
c
4
x
2
+
2
c
5
x
5
−
3
(
5
)
c
5
x
4
+
5
(
4
)
c
5
x
3
=
4
x
2
−
6
x
5
{\displaystyle \ 2c_{0}+2c_{1}x-3c_{1}+2c_{2}x^{2}-3c_{2}(2)x+2c_{2}+2c_{3}x^{3}-3(3)c_{3}x^{2}+3(2)c_{3}x+2c_{4}x^{4}-3(4)c_{4}x^{3}+4(3)c_{4}x^{2}+2c_{5}x^{5}-3(5)c_{5}x^{4}+5(4)c_{5}x^{3}=4x^{2}-6x^{5}\ }
Simplifying and grouping like terms we get:
(
2
c
0
−
3
c
1
+
2
c
2
)
+
(
2
c
1
−
6
c
2
+
6
c
3
)
x
+
(
2
c
2
−
9
c
3
+
12
c
4
)
x
2
+
(
2
c
3
−
12
c
4
+
20
c
5
)
x
3
+
(
2
c
4
−
15
c
5
)
x
4
+
2
c
5
x
5
=
4
x
2
−
6
x
5
{\displaystyle \ (2c_{0}-3c_{1}+2c_{2})+(2c_{1}-6c_{2}+6c_{3})x+(2c_{2}-9c_{3}+12c_{4})x^{2}+(2c_{3}-12c_{4}+20c_{5})x^{3}+(2c_{4}-15c_{5})x^{4}+2c_{5}x^{5}=4x^{2}-6x^{5}\ }
Compared to the answers of part 1, it can be seen that all the coefficients are equivalent, verifying that the two series are equivalent.
Matrix form of the solution:
[
2
−
3
2
0
0
0
0
2
−
6
6
0
0
0
0
2
−
9
12
0
0
0
0
2
−
12
20
0
0
0
0
2
−
15
0
0
0
0
0
2
]
[
c
0
c
1
c
2
c
3
c
4
c
5
]
=
[
0
0
4
0
0
−
6
]
{\displaystyle \ {\begin{bmatrix}2&-3&2&0&0&0\\0&2&-6&6&0&0\\0&0&2&-9&12&0\\0&0&0&2&-12&20\\0&0&0&0&2&-15\\0&0&0&0&0&2\end{bmatrix}}{\begin{bmatrix}c_{0}\\c_{1}\\c_{2}\\c_{3}\\c_{4}\\c_{5}\end{bmatrix}}={\begin{bmatrix}0\\0\\4\\0\\0\\-6\end{bmatrix}}\ }
Solving by back substitution, the first constant we solve for is c5
2
c
5
=
−
6
{\displaystyle \ 2c_{5}=-6\ }
c
5
=
−
3
{\displaystyle \ c_{5}=-3\ }
−
15
c
5
+
2
c
4
=
0
{\displaystyle \ -15c_{5}+2c_{4}=0\ }
−
15
(
−
3
)
+
2
c
4
=
0
{\displaystyle \ -15(-3)+2c_{4}=0\ }
c
4
=
−
22.5
{\displaystyle \ c_{4}=-22.5\ }
20
c
5
−
12
c
4
+
2
c
3
=
0
{\displaystyle \ 20c_{5}-12c_{4}+2c_{3}=0\ }
20
(
−
3
)
−
12
(
−
22.5
)
+
2
c
3
=
0
{\displaystyle \ 20(-3)-12(-22.5)+2c_{3}=0\ }
c
3
=
−
105
{\displaystyle \ c_{3}=-105\ }
12
c
4
−
9
c
3
+
2
c
2
=
4
{\displaystyle \ 12c_{4}-9c_{3}+2c_{2}=4\ }
12
(
−
22.5
)
−
9
(
−
105
)
+
2
c
2
=
4
{\displaystyle \ 12(-22.5)-9(-105)+2c_{2}=4\ }
c
2
=
−
335.5
{\displaystyle \ c_{2}=-335.5\ }
6
c
3
−
6
c
2
+
2
c
1
=
0
{\displaystyle \ 6c_{3}-6c_{2}+2c_{1}=0\ }
6
(
−
105
)
−
6
(
−
335.5
)
+
2
c
1
=
0
{\displaystyle \ 6(-105)-6(-335.5)+2c_{1}=0\ }
c
1
=
−
691.5
{\displaystyle \ c_{1}=-691.5\ }
2
c
2
−
3
c
1
+
2
c
0
=
0
{\displaystyle \ 2c_{2}-3c_{1}+2c_{0}=0\ }
2
(
−
335.5
)
−
3
(
−
691.5
)
+
2
c
0
=
0
{\displaystyle \ 2(-335.5)-3(-691.5)+2c_{0}=0\ }
c
0
=
−
701.75
{\displaystyle c_{0}=-701.75\ }
The particular solution is of the form
y
p
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
{\displaystyle \ y_{p}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\ }
Therefore, the particular solution to the differential equation =
y
p
=
−
701.75
−
691.5
x
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
{\displaystyle \ y_{p}=-701.75-691.5x-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}\ }
To find the general solution, we need to add the particular solution to the homogenous solution.
The homogenous solution is given by:
(
λ
1
−
1
)
(
λ
2
−
2
)
=
0
{\displaystyle \ (\lambda _{1}-1)(\lambda _{2}-2)=0\ }
y
h
=
C
1
e
λ
1
x
+
C
2
e
λ
2
x
=
C
1
e
x
+
C
2
e
2
x
{\displaystyle \ y_{h}=C_{1}e^{\lambda _{1}x}+C_{2}e^{\lambda _{2}}x=C_{1}e^{x}+C_{2}e^{2x}\ }
Therefore the general solution is:
y
h
+
y
p
=
C
1
e
x
+
C
2
e
2
x
−
701.75
−
691.5
x
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
{\displaystyle \ y_{h}+y_{p}=C_{1}e^{x}+C_{2}e^{2x}-701.75-691.5x-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}\ }
Now using the initial conditions to solve for the remaining constants we need the general solution and the derivative of the general solution.
y
′
(
x
)
=
C
1
e
x
+
2
C
2
e
2
x
−
691.5
−
671
x
−
315
x
2
−
90
x
3
−
15
x
4
{\displaystyle \ y'(x)=C_{1}e^{x}+2C_{2}e^{2x}-691.5-671x-315x^{2}-90x^{3}-15x^{4}\ }
y
(
0
)
=
1
=
C
1
+
C
2
−
701.75
{\displaystyle \ y(0)=1=C_{1}+C_{2}-701.75\ }
y
′
(
0
)
=
0
=
C
1
+
2
C
2
−
691.5
{\displaystyle \ y'(0)=0=C_{1}+2C_{2}-691.5\ }
We can move the 1 over, and then subtract the equations to solve for
C
2
{\displaystyle \ C_{2}\ }
−
C
2
−
11.25
=
0
⇒
C
2
=
−
11.25
{\displaystyle \ -C_{2}-11.25=0\Rightarrow C_{2}=-11.25\ }
C
1
+
C
2
−
701.75
=
1
{\displaystyle \ C_{1}+C_{2}-701.75=1\ }
C
1
=
1
+
701.75
−
−
11.25
=
714
{\displaystyle \ C_{1}=1+701.75--11.25=714\ }
Therefore the final general solution y(x) is:
y
(
x
)
=
714
e
x
−
11.25
e
2
x
−
701.75
−
691.5
x
−
335.5
x
2
−
105
x
3
−
22.5
x
4
−
3
x
5
{\displaystyle \ y(x)=714e^{x}-11.25e^{2x}-701.75-691.5x-335.5x^{2}-105x^{3}-22.5x^{4}-3x^{5}\ }
Matlab code:
x = 0:0.001:100;
y = 714*exp(x) - 11.25*exp(2*x) - 701.75 - 691.5*x - 335.5*x.^2 - 105*x.^3 - 22.5*x.^4 - 3*x.^5;
plot(x,y)
xlabel('x')
ylabel('y(x)')
This problem was solved and uploaded by: David Herrick
This problem was proofread by Michael Wallace
Solve the L2-ODE-CC (2) p.7-11 with initial conditions (2b) p.3-6 differently as follows. Consider the following two L2-ODE-CC (see p. 7-2b)
y
p
,
1
″
−
3
y
p
,
1
′
+
2
y
p
,
1
=
r
1
(
x
)
=
4
x
2
{\displaystyle y_{p,1}''-3y_{p,1}'+2y_{p,1}=r_{1}(x)=4x^{2}\!}
y
p
,
2
″
−
3
y
p
,
2
′
+
2
y
p
,
2
=
r
2
(
x
)
=
−
6
x
5
{\displaystyle y_{p,2}''-3y_{p,2}'+2y_{p,2}=r_{2}(x)=-6x^{5}\!}
The particular solution to
y
p
,
1
{\displaystyle y_{p,1}\!}
Find the particular solution
y
p
,
2
{\displaystyle y_{p,2}\!}
y
{\displaystyle y\!}
First particular solution:
y
p
,
1
=
2
x
2
+
6
x
+
7
{\displaystyle y_{p,1}=2x^{2}+6x+7\!}
Initial Conditions:
y
(
0
)
=
1
{\displaystyle y(0)=1\!}
y
′
(
0
)
=
0
{\displaystyle y'(0)=0\!}
Because of the specific excitation
r
2
(
x
)
=
−
6
x
5
{\displaystyle r_{2}(x)=-6x^{5}\!}
y
p
(
x
)
=
∑
j
=
0
n
K
j
x
j
{\displaystyle y_{p}(x)=\sum _{j=0}^{n}K_{j}x^{j}\!}
Then the following is presented:
y
p
,
2
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
{\displaystyle y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!}
y
p
,
2
′
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
{\displaystyle y_{p,2}'=c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}\!}
y
p
,
2
″
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
{\displaystyle y_{p,2}''=2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}\!}
Plug these equations back into the original
y
p
,
2
″
−
3
y
p
,
2
′
+
2
y
p
,
2
=
−
6
x
5
{\displaystyle y_{p,2}''-3y_{p,2}'+2y_{p,2}=-6x^{5}\!}
(
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
)
−
3
(
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
)
+
2
(
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
)
=
−
6
x
5
{\displaystyle (2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3})-3(c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4})+2(c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5})=-6x^{5}\!}
Use coefficient matching, the following equations are a result:
c
:
2
c
2
−
3
c
1
+
2
c
0
=
0
{\displaystyle c:2c_{2}-3c_{1}+2c_{0}=0\!}
x
:
6
c
3
−
6
c
2
+
2
c
1
=
0
{\displaystyle x:6c_{3}-6c_{2}+2c_{1}=0\!}
x
2
:
12
c
4
−
9
c
3
+
2
c
2
=
0
{\displaystyle x^{2}:12c_{4}-9c_{3}+2c_{2}=0\!}
x
3
:
20
c
5
−
12
c
4
+
2
c
3
=
0
{\displaystyle x^{3}:20c_{5}-12c_{4}+2c_{3}=0\!}
x
4
:
−
15
c
5
+
2
c
4
=
0
{\displaystyle x^{4}:-15c_{5}+2c_{4}=0\!}
x
5
:
2
c
5
=
−
6
{\displaystyle x^{5}:2c_{5}=-6\!}
Use back substitution method to solve for every coefficient, starting with
c
5
{\displaystyle c_{5}\!}
2
c
5
=
−
6
→
c
5
=
−
3
{\displaystyle 2c_{5}=-6\rightarrow c_{5}=-3\!}
2
c
4
−
15
(
−
3
)
=
0
→
c
4
=
−
45
2
{\displaystyle 2c_{4}-15(-3)=0\rightarrow c_{4}=-{\frac {45}{2}}\!}
2
c
3
−
12
(
−
45
2
)
+
20
(
−
3
)
=
0
→
c
3
=
−
105
{\displaystyle 2c_{3}-12(-{\frac {45}{2}})+20(-3)=0\rightarrow c_{3}=-105\!}
2
c
2
−
9
(
−
105
)
+
12
(
−
45
2
)
=
0
→
c
2
=
−
675
2
{\displaystyle 2c_{2}-9(-105)+12(-{\frac {45}{2}})=0\rightarrow c_{2}=-{\frac {675}{2}}\!}
2
c
1
−
6
(
−
675
2
)
+
6
(
−
105
)
=
0
→
c
1
=
−
1395
2
{\displaystyle 2c_{1}-6(-{\frac {675}{2}})+6(-105)=0\rightarrow c_{1}=-{\frac {1395}{2}}\!}
2
c
0
−
3
(
−
1395
2
)
+
2
(
−
675
2
)
=
0
→
c
0
=
−
2835
4
{\displaystyle 2c_{0}-3(-{\frac {1395}{2}})+2(-{\frac {675}{2}})=0\rightarrow c_{0}=-{\frac {2835}{4}}\!}
Plug these values back into
y
p
,
2
=
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
+
c
5
x
5
{\displaystyle y_{p,2}=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}\!}
y
p
,
2
=
−
2835
4
−
1395
2
x
−
675
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y_{p,2}=-{\frac {2835}{4}}-{\frac {1395}{2}}x-{\frac {675}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
Superposition principle applies ( L2-ODE-CC ),
y
p
=
y
p
,
1
+
y
p
,
2
{\displaystyle y_{p}=y_{p,1}+y_{p,2}\!}
y
p
(
x
)
=
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y_{p}(x)=-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
Homogeneous Solution
edit
y
h
″
−
3
y
h
′
+
2
y
h
=
0
{\displaystyle y_{h}''-3y_{h}'+2y_{h}=0\!}
Because of the linearity, a combination of two linear independent solutions is also a solution to this homogeneous equation:
y
h
,
1
″
−
3
y
h
,
1
′
+
2
y
h
,
1
=
0
{\displaystyle y_{h,1}''-3y_{h,1}'+2y_{h,1}=0\!}
y
h
,
2
″
−
3
y
h
,
2
′
+
2
y
h
,
2
=
0
{\displaystyle y_{h,2}''-3y_{h,2}'+2y_{h,2}=0\!}
With the solutions as:
y
h
,
1
=
e
λ
1
x
{\displaystyle y_{h,1}=e^{\lambda _{1}x}\!}
y
h
,
2
=
e
λ
2
x
{\displaystyle y_{h,2}=e^{\lambda _{2}x}\!}
To determine the value of
λ
1
,
2
{\displaystyle \lambda _{1,2}\!}
λ
2
−
3
λ
+
2
=
0
{\displaystyle \lambda ^{2}-3\lambda +2=0\!}
λ
1
,
2
=
−
a
±
a
2
−
4
b
2
{\displaystyle \lambda _{1,2}={\frac {-a\pm {\sqrt {a^{2}-4b}}}{2}}\!}
λ
1
,
2
=
−
(
−
3
)
±
(
−
3
)
2
−
4
(
2
)
2
{\displaystyle \lambda _{1,2}={\frac {-(-3)\pm {\sqrt {(-3)^{2}-4(2)}}}{2}}\!}
λ
1
=
2
λ
2
=
1
{\displaystyle \lambda _{1}=2\;\;\;\;\;\;\;\lambda _{2}=1\!}
The solutions for each distinct linearly independent homogeneous equation become:
y
h
,
1
=
e
2
x
{\displaystyle y_{h,1}=e^{2x}\!}
y
h
,
2
=
e
x
{\displaystyle y_{h,2}=e^{x}\!}
The combination of the previous two equations, multiplied by two constants that satisfy two initial conditions, is also a solution:
y
h
(
x
)
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle y_{h}(x)=C_{1}e^{2x}+C_{2}e^{x}\!}
The general or overall solution for the L2-ODE-CC:
y
(
x
)
=
y
h
+
y
p
{\displaystyle y(x)=y_{h}+y_{p}\!}
y
(
x
)
=
C
1
e
2
x
+
C
2
e
x
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y(x)=C_{1}e^{2x}+C_{2}e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
y
′
(
x
)
=
2
C
1
e
2
x
+
C
2
e
x
−
1383
2
−
671
x
−
315
x
2
−
90
x
3
−
15
x
4
{\displaystyle y'(x)=2C_{1}e^{2x}+C_{2}e^{x}-{\frac {1383}{2}}-671x-315x^{2}-90x^{3}-15x^{4}\!}
Use initial conditions to solve for
C
1
{\displaystyle C_{1}\!}
C
2
{\displaystyle C_{2}\!}
1
=
C
1
+
C
2
−
2807
4
{\displaystyle 1=C_{1}+C_{2}-{\frac {2807}{4}}\!}
0
=
C
1
+
C
2
−
1383
2
{\displaystyle 0=C_{1}+C_{2}-{\frac {1383}{2}}\!}
C
1
=
−
45
4
C
2
=
714
{\displaystyle C_{1}=-{\frac {45}{4}}\;\;\;C_{2}=714\!}
The general solution:
y
(
x
)
=
−
45
4
e
2
x
+
714
e
x
−
2807
4
−
1383
2
x
−
671
2
x
2
−
105
x
3
−
45
2
x
4
−
3
x
5
{\displaystyle y(x)=-{\frac {45}{4}}e^{2x}+714e^{x}-{\frac {2807}{4}}-{\frac {1383}{2}}x-{\frac {671}{2}}x^{2}-105x^{3}-{\frac {45}{2}}x^{4}-3x^{5}\!}
This problem was solved and uploaded by Derik Bell
This problem was proofread by Michael Wallace
Expand the series on both sides of (1)-(2) p.7-12b to verify these equalities.
Equation (1)
∑
j
=
2
5
c
j
j
(
j
−
1
)
x
j
−
2
=
∑
j
=
0
3
c
j
+
2
(
j
+
2
)
(
j
+
1
)
x
j
{\displaystyle \displaystyle \sum _{j=2}^{5}c_{j}j(j-1)x^{j-2}=\sum _{j=0}^{3}c_{j+2}(j+2)(j+1)x^{j}}
Equation (2)
∑
j
=
1
5
c
j
j
x
j
−
1
=
∑
j
=
0
4
c
j
+
1
(
j
+
1
)
x
j
{\displaystyle \displaystyle \sum _{j=1}^{5}c_{j}jx^{j-1}=\sum _{j=0}^{4}c_{j+1}(j+1)x^{j}}
Expanding both sides of Equation (1) yields the following expression:
c
2
(
2
)
(
1
)
x
0
+
c
3
(
3
)
(
2
)
x
+
c
4
(
4
)
(
3
)
x
2
+
c
5
(
5
)
(
4
)
x
3
=
2
c
2
+
6
c
3
x
+
12
c
4
x
2
+
20
c
5
x
3
{\displaystyle \displaystyle c_{2}(2)(1)x^{0}+c_{3}(3)(2)x+c_{4}(4)(3)x^{2}+c_{5}(5)(4)x^{3}=2c_{2}+6c_{3}x+12c_{4}x^{2}+20c_{5}x^{3}}
Therefore, they are equivalent.
Expanding both sides of Equation (2) yields the following expression:
c
1
(
1
)
x
0
+
c
2
(
2
)
x
1
+
c
3
(
3
)
x
2
+
c
4
(
4
)
x
3
+
c
5
(
5
)
x
4
=
c
1
+
2
c
2
x
+
3
c
3
x
2
+
4
c
4
x
3
+
5
c
5
x
4
{\displaystyle \displaystyle c_{1}(1)x^{0}+c_{2}(2)x^{1}+c_{3}(3)x^{2}+c_{4}(4)x^{3}+c_{5}(5)x^{4}=c_{1}+2c_{2}x+3c_{3}x^{2}+4c_{4}x^{3}+5c_{5}x^{4}}
Therefore, they are equivalent.
This problem was solved and uploaded by: William Knapper
This problem was proofread by: Joshua House
Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.
8
y
″
−
6
y
′
+
y
=
6
c
o
s
h
x
,
y
(
0
)
=
0.2
,
y
′
(
0
)
=
0.05
{\displaystyle \displaystyle 8y''-6y'+y=6coshx,y(0)=0.2,y'(0)=0.05}
y
″
+
4
y
′
+
4
y
=
e
−
2
x
s
i
n
2
x
,
y
(
0
)
=
1
,
y
′
(
0
)
=
−
1.5
{\displaystyle \displaystyle y''+4y'+4y=e^{-2x}sin2x,y(0)=1,y'(0)=-1.5}
First, we find the homogeneous ODE solution
8
λ
2
−
6
λ
+
1
=
0
{\displaystyle \displaystyle 8\lambda ^{2}-6\lambda +1=0}
(
4
λ
−
1
)
(
2
λ
−
1
)
=
0
{\displaystyle \displaystyle (4\lambda -1)(2\lambda -1)=0}
λ
=
1
4
,
λ
=
1
2
{\displaystyle \displaystyle \lambda ={\frac {1}{4}},\lambda ={\frac {1}{2}}}
which gives the general solution:
y
h
=
c
1
e
1
4
x
+
c
2
e
1
2
x
{\displaystyle \displaystyle y_{h}=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}}
Second, we get the particular solution of the non homogeneous ODE.
c
o
s
h
x
=
e
x
+
e
−
x
2
{\displaystyle coshx={\frac {e^{x}+e^{-x}}{2}}}
y
p
=
y
p
1
+
y
p
2
{\displaystyle y_{p}=y_{p1}+y_{p2}}
y
p
1
=
c
3
e
x
,
y
p
2
=
c
4
e
−
x
{\displaystyle \displaystyle y_{p1}=c_{3}e^{x},y_{p2}=c_{4}e^{-x}}
Then we get
y
p
=
c
3
e
x
+
c
4
e
−
x
{\displaystyle \displaystyle y_{p}=c_{3}e^{x}+c_{4}e^{-x}}
y
p
′
=
c
3
e
x
−
c
4
e
−
x
{\displaystyle \displaystyle y_{p}'=c_{3}e^{x}-c_{4}e^{-x}}
y
p
″
=
c
3
e
x
+
c
4
e
−
x
{\displaystyle \displaystyle y_{p}''=c_{3}e^{x}+c_{4}e^{-x}}
The original ODE becomes:
8
y
″
−
6
y
′
+
y
=
3
e
x
+
3
e
−
x
{\displaystyle \displaystyle 8y''-6y'+y=3e^{x}+3e^{-x}}
Substituting
y
p
{\displaystyle y_{p}}
8
c
3
−
6
c
3
+
c
3
=
3
{\displaystyle \displaystyle 8c_{3}-6c_{3}+c_{3}=3}
and
8
c
4
+
6
c
4
+
c
4
=
3
{\displaystyle \displaystyle 8c_{4}+6c_{4}+c_{4}=3}
Solving these two equations we get:
c
3
=
1
{\displaystyle \displaystyle c_{3}=1}
c
4
=
1
5
{\displaystyle \displaystyle c_{4}={\frac {1}{5}}}
Thus,
y
p
=
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle y_{p}=e^{x}+{\frac {1}{5}}e^{-x}}
We know that
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}}
y
=
c
1
e
1
4
x
+
c
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle y=c_{1}e^{{\frac {1}{4}}x}+c_{2}e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}
Finding the solution to the initial value problem:
y
(
0
)
=
0.2
=
c
1
(
1
)
+
c
2
(
1
)
+
1
+
1
5
{\displaystyle \displaystyle y(0)=0.2=c_{1}(1)+c_{2}(1)+1+{\frac {1}{5}}}
−
1
=
c
1
+
c
2
{\displaystyle \displaystyle -1=c_{1}+c_{2}}
c
2
=
−
c
1
−
1
{\displaystyle \displaystyle c_{2}=-c_{1}-1}
y
′
=
1
4
c
1
e
1
4
x
+
1
2
c
2
e
1
2
x
+
e
x
−
1
5
e
−
x
{\displaystyle \displaystyle y'={\frac {1}{4}}c_{1}e^{{\frac {1}{4}}x}+{\frac {1}{2}}c_{2}e^{{\frac {1}{2}}x}+e^{x}-{\frac {1}{5}}e^{-x}}
y
′
(
0
)
=
0.05
=
1
4
c
1
(
1
)
+
1
2
c
2
(
1
)
+
1
−
1
5
{\displaystyle \displaystyle y'(0)=0.05={\frac {1}{4}}c_{1}(1)+{\frac {1}{2}}c_{2}(1)+1-{\frac {1}{5}}}
−
0.75
=
1
4
c
1
+
1
2
c
2
{\displaystyle \displaystyle -0.75={\frac {1}{4}}c_{1}+{\frac {1}{2}}c_{2}}
Substituting the known value for
c
2
{\displaystyle c_{2}}
−
0.75
=
1
4
c
1
+
1
2
(
−
c
1
−
1
)
{\displaystyle \displaystyle -0.75={\frac {1}{4}}c_{1}+{\frac {1}{2}}(-c_{1}-1)}
−
0.75
=
1
4
c
1
−
1
2
c
1
−
1
2
{\displaystyle \displaystyle -0.75={\frac {1}{4}}c_{1}-{\frac {1}{2}}c_{1}-{\frac {1}{2}}}
−
0.25
=
−
1
4
c
1
{\displaystyle \displaystyle -0.25=-{\frac {1}{4}}c_{1}}
Solving this we find that:
c
1
=
1
{\displaystyle \displaystyle c_{1}=1}
c
2
=
−
2
{\displaystyle \displaystyle c_{2}=-2}
The final solution is:
y
=
e
1
4
x
−
2
e
1
2
x
+
e
x
+
1
5
e
−
x
{\displaystyle \displaystyle y=e^{{\frac {1}{4}}x}-2e^{{\frac {1}{2}}x}+e^{x}+{\frac {1}{5}}e^{-x}}
First, we find the homogeneous ODE solution
λ
2
+
4
λ
+
4
=
0
{\displaystyle \displaystyle \lambda ^{2}+4\lambda +4=0}
(
λ
+
2
)
(
λ
+
2
)
=
0
{\displaystyle \displaystyle (\lambda +2)(\lambda +2)=0}
λ
=
−
2
,
λ
=
−
2
{\displaystyle \displaystyle \lambda =-2,\lambda =-2}
which gives the general solution:
y
h
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
{\displaystyle \displaystyle y_{h}=c_{1}e^{-2x}+c_{2}xe^{-2x}}
Second, we get the particular solution of the non homogeneous ODE.
y
p
=
e
−
2
x
(
K
c
o
s
2
x
+
M
s
i
n
2
x
)
{\displaystyle \displaystyle y_{p}=e^{-2x}(Kcos2x+Msin2x)}
e
−
2
x
{\displaystyle e^{-2x}}
y
p
{\displaystyle y_{p}}
y
p
=
x
e
−
2
x
(
K
c
o
s
2
x
+
M
s
i
n
2
x
)
{\displaystyle \displaystyle y_{p}=xe^{-2x}(Kcos2x+Msin2x)}
Then we find:
y
p
′
=
(
e
−
2
x
−
2
x
e
−
2
x
)
(
K
c
o
s
2
x
+
M
s
i
n
2
x
)
+
x
e
−
2
x
(
−
2
K
s
i
n
2
x
+
2
M
c
o
s
2
x
)
{\displaystyle \displaystyle y_{p}'=(e^{-2x}-2xe^{-2x})(Kcos2x+Msin2x)+xe^{-2x}(-2Ksin2x+2Mcos2x)}
y
p
′
=
K
e
−
2
x
c
o
s
2
x
−
2
K
x
e
−
2
x
c
o
s
2
x
+
M
e
−
2
x
s
i
n
2
x
−
2
M
x
e
−
2
x
s
i
n
2
x
−
2
K
x
e
−
2
x
s
i
n
2
x
+
2
M
x
e
−
2
x
c
o
x
2
x
{\displaystyle \displaystyle y_{p}'=Ke^{-2x}cos2x-2Kxe^{-2x}cos2x+Me^{-2x}sin2x-2Mxe^{-2x}sin2x-2Kxe^{-2x}sin2x+2Mxe^{-2x}cox2x}
y
p
′
=
K
e
−
2
x
c
o
s
2
x
+
M
e
−
2
x
s
i
n
2
x
+
(
−
2
K
−
2
M
)
x
e
−
2
x
s
i
n
2
x
+
(
−
2
K
+
2
M
)
x
e
−
2
x
c
o
s
2
x
{\displaystyle \displaystyle y_{p}'=Ke^{-2x}cos2x+Me^{-2x}sin2x+(-2K-2M)xe^{-2x}sin2x+(-2K+2M)xe^{-2x}cos2x}
And,
y
p
″
=
K
[
−
2
e
−
2
x
c
o
s
2
x
−
2
e
−
2
x
s
i
n
2
x
]
+
M
[
−
2
e
−
2
x
s
i
n
2
x
+
2
e
−
2
x
c
o
s
2
x
]
+
(
−
2
K
+
2
M
)
[
(
e
−
2
x
−
2
x
e
−
2
x
)
(
c
o
s
2
x
)
+
(
−
2
x
e
−
2
x
s
i
n
2
x
)
]
{\displaystyle \displaystyle y_{p}''=K[-2e^{-2x}cos2x-2e^{-2x}sin2x]+M[-2e^{-2x}sin2x+2e^{-2x}cos2x]+(-2K+2M)[(e^{-2x}-2xe^{-2x})(cos2x)+(-2xe^{-2x}sin2x)]}
+
(
−
2
K
−
2
M
)
[
(
e
−
2
x
−
2
x
e
−
2
x
)
(
s
i
n
2
x
)
+
(
2
x
e
−
2
x
c
o
s
2
x
)
]
{\displaystyle \displaystyle +(-2K-2M)[(e^{-2x}-2xe^{-2x})(sin2x)+(2xe^{-2x}cos2x)]}
y
p
″
=
−
2
K
e
−
2
x
c
o
s
2
x
−
2
K
e
−
2
x
s
i
n
2
x
−
M
2
e
−
2
x
s
i
n
2
x
+
2
M
e
−
2
x
c
o
s
2
x
−
2
K
e
−
2
x
c
o
s
2
x
+
4
K
x
e
−
2
x
s
i
n
2
x
+
4
K
x
e
−
2
x
c
o
s
2
x
{\displaystyle \displaystyle y_{p}''=-2Ke^{-2x}cos2x-2Ke^{-2x}sin2x-M2e^{-2x}sin2x+2Me^{-2x}cos2x-2Ke^{-2x}cos2x+4Kxe^{-2x}sin2x+4Kxe^{-2x}cos2x}
+
2
M
e
−
2
x
c
o
s
2
x
−
4
M
x
e
−
2
x
c
o
s
2
x
−
4
M
x
e
−
2
x
s
i
n
2
x
−
2
K
e
−
2
x
s
i
n
2
x
+
4
K
x
e
−
2
x
s
i
n
2
x
−
4
K
x
e
−
2
x
c
o
s
2
x
{\displaystyle \displaystyle +2Me^{-2x}cos2x-4Mxe{^{-}2x}cos2x-4Mxe^{-2x}sin2x-2Ke^{-2x}sin2x+4Kxe^{-2x}sin2x-4Kxe^{-2x}cos2x}
−
2
M
e
−
2
x
s
i
n
2
x
+
4
M
x
e
−
2
x
s
i
n
2
x
−
4
M
x
e
−
2
x
c
o
s
2
x
{\displaystyle \displaystyle -2Me^{-2x}sin2x+4Mxe^{-2x}sin2x-4Mxe^{-2x}cos2x}
y
p
″
=
(
−
4
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
(
−
4
K
−
4
M
)
e
−
2
x
s
i
n
2
x
+
8
K
x
e
−
2
x
s
i
n
2
x
−
8
M
x
e
−
2
x
c
o
s
2
x
{\displaystyle \displaystyle y_{p}''=(-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x+8Kxe^{-2x}sin2x-8Mxe^{-2x}cos2x}
Substituting
y
p
{\displaystyle y_{p}}
(
−
4
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
(
−
4
K
−
4
M
)
e
−
2
x
s
i
n
2
x
+
8
K
x
e
−
2
x
s
i
n
2
x
−
8
M
x
e
−
2
x
c
o
s
2
x
−
4
K
e
−
2
x
c
o
s
2
x
+
4
M
e
−
2
x
s
i
n
2
x
{\displaystyle \displaystyle (-4K+4M)e^{-2x}cos2x+(-4K-4M)e^{-2x}sin2x+8Kxe^{-2x}sin2x-8Mxe^{-2x}cos2x-4Ke^{-2x}cos2x+4Me^{-2x}sin2x}
+
(
−
8
K
−
8
M
)
x
e
−
2
x
s
i
n
2
x
+
(
−
8
K
+
8
M
)
x
e
−
2
x
c
o
s
2
x
+
4
K
x
e
−
2
x
c
o
s
2
x
+
4
M
x
e
−
2
x
s
i
n
2
x
=
e
−
2
x
s
i
n
2
x
{\displaystyle \displaystyle +(-8K-8M)xe^{-2x}sin2x+(-8K+8M)xe^{-2x}cos2x+4Kxe^{-2x}cos2x+4Mxe^{-2x}sin2x=e^{-2x}sin2x}
(
−
8
K
+
4
M
)
e
−
2
x
c
o
s
2
x
+
−
4
K
e
−
2
x
s
i
n
2
x
=
e
−
2
x
s
i
n
2
x
{\displaystyle \displaystyle (-8K+4M)e^{-2x}cos2x+-4Ke^{-2x}sin2x=e^{-2x}sin2x}
Separating into components we get:
−
4
K
=
1
{\displaystyle \displaystyle -4K=1}
−
8
K
+
4
M
=
0
{\displaystyle \displaystyle -8K+4M=0}
Solving the two equations we get:
K
=
−
1
4
{\displaystyle \displaystyle K=-{\frac {1}{4}}}
M
=
−
1
2
{\displaystyle \displaystyle M=-{\frac {1}{2}}}
Thus,
y
p
=
e
−
2
x
(
−
1
4
c
o
s
2
x
−
1
2
s
i
n
2
x
)
{\displaystyle \displaystyle y_{p}=e^{-2x}(-{\frac {1}{4}}cos2x-{\frac {1}{2}}sin2x)}
We know that
y
=
y
h
+
y
p
{\displaystyle y=y_{h}+y_{p}}
y
=
c
1
e
−
2
x
+
c
2
x
e
−
2
x
+
e
−
2
x
(
−
1
4
c
o
s
2
x
−
1
2
s
i
n
2
x
)
{\displaystyle \displaystyle y=c_{1}e^{-2x}+c_{2}xe^{-2x}+e^{-2x}(-{\frac {1}{4}}cos2x-{\frac {1}{2}}sin2x)}
Finding the solution to the initial value problem:
y
(
0
)
=
1
=
c
1
(
1
)
+
(
1
)
[
−
1
4
(
1
)
]
{\displaystyle \displaystyle y(0)=1=c_{1}(1)+(1)[-{\frac {1}{4}}(1)]}
We solve,
c
1
=
−
5
4
{\displaystyle \displaystyle c_{1}=-{\frac {5}{4}}}
Also,
y
′
=
−
2
c
1
e
−
2
x
−
2
c
2
x
e
−
2
x
+
c
2
e
−
2
x
−
2
e
−
2
x
(
−
1
4
c
o
s
2
x
−
1
2
s
i
n
2
x
)
+
e
−
2
x
(
1
2
s
i
n
2
x
−
c
o
s
2
x
)
{\displaystyle \displaystyle y'=-2c_{1}e^{-2x}-2c_{2}xe^{-2x}+c_{2}e^{-2x}-2e^{-2x}(-{\frac {1}{4}}cos2x-{\frac {1}{2}}sin2x)+e^{-2x}({\frac {1}{2}}sin2x-cos2x)}
y
′
(
0
)
=
−
1.5
=
−
2
(
−
5
4
)
(
1
)
+
c
2
(
1
)
−
2
(
1
)
[
−
1
4
(
1
)
]
+
(
1
)
(
−
1
)
{\displaystyle \displaystyle y'(0)=-1.5=-2(-{\frac {5}{4}})(1)+c_{2}(1)-2(1)[-{\frac {1}{4}}(1)]+(1)(-1)}
We solve,
c
2
=
−
3.5
=
−
7
2
{\displaystyle \displaystyle c_{2}=-3.5=-{\frac {7}{2}}}
The final solution is:
y
=
−
5
4
e
−
2
x
−
7
2
x
e
−
2
x
+
e
−
2
x
(
−
1
4
c
o
s
2
x
−
1
2
s
i
n
2
x
)
{\displaystyle \displaystyle y=-{\frac {5}{4}}e^{-2x}-{\frac {7}{2}}xe^{-2x}+e^{-2x}(-{\frac {1}{4}}cos2x-{\frac {1}{2}}sin2x)}
This problem was solved and uploaded by: Radina Dikova
This problem was proofread by: David Herrick
Contribution Summary
edit
![{\displaystyle \displaystyle C[2+10x+10x+25x^{2}-20x-50x^{2}+25x^{2}]e^{5x}=7e^{5x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fa6cfe5ef82e4b3ad556f9357849497bf5aad08)
![{\displaystyle \displaystyle y^{'}=e^{5x}[5c_{1}+c_{2}+5c_{2}x+7x+{\frac {35}{2}}x^{2}]-{\frac {4}{25}}x-{\frac {8}{125}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9dcb7f4a676445f9f0b73fd9201d12abec45b6f)
![{\displaystyle \displaystyle (\lambda -x)(\lambda -[x+\varepsilon ])=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8a3ae471f10c215ef3a4ed19c15999e708695092)
![{\displaystyle \displaystyle \lambda ^{2}-\lambda [2x+\varepsilon ]+x[x+\varepsilon ]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89f1091e339382cd55bce680de06881f9ec6dd9a)
![{\displaystyle \displaystyle y^{''}-y^{'}[2\lambda +\varepsilon ]+y\lambda [\lambda +\varepsilon ]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f085fa467dca5572bdd4808d14b795fe6df7af1)
![{\displaystyle \displaystyle e^{(\lambda +\varepsilon )x}[\lambda ^{2}+2\lambda \varepsilon +\varepsilon ^{2}-2\lambda ^{2}-2\lambda \varepsilon -\lambda \varepsilon -\varepsilon ^{2}+\lambda ^{2}+\varepsilon \lambda ]+e^{\lambda x}[-\lambda ^{2}+2\lambda ^{2}+\lambda \varepsilon -\lambda ^{2}-\lambda \varepsilon ]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d93264a149b0d80b4c076019ce3a4d0f74a31486)
![{\displaystyle \displaystyle {\frac {d[e^{\lambda x}]}{d\lambda }}=xe^{\lambda x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91c24297f4d2038b2e809ef7359f687d28f098dd)
![{\displaystyle \displaystyle {\frac {d[e^{\lambda x}]}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\varepsilon )x}-e^{\lambda x}}{\varepsilon }}=xe^{\lambda x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46f210a0de3c1180b1d0c4305c20b65e94ff8307)
![{\displaystyle \displaystyle e^{5.001x}[5^{2}+2(5)(0.001)+0.001^{2}-2(5^{2})-2(5)(0.001)-5(0.001)-(0.001^{2})+5^{2}+0.001(5)]+e^{5x}[-5^{2}+2(5^{2})+5(0.001)-5^{2}-5(0.001)]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bea4ac5f40767e8be2940b165c9853a87482abce)
![{\displaystyle \ \sum _{j=0}^{3}\left[c_{j+2}(j+2)(j+1)-3c_{j+1}(j+1)+2c_{j}\right]x^{j}-3c_{5}(5)x^{4}+2\left[c_{4}x^{4}+c_{5}x^{5}\right]=4x^{2}-6x^{5}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e39e974c470ba6b22e5011da08601bdfeca549a3)
![{\displaystyle \displaystyle y_{p}''=K[-2e^{-2x}cos2x-2e^{-2x}sin2x]+M[-2e^{-2x}sin2x+2e^{-2x}cos2x]+(-2K+2M)[(e^{-2x}-2xe^{-2x})(cos2x)+(-2xe^{-2x}sin2x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8730336f5435750b3c138c9952c7228b61e0be94)
![{\displaystyle \displaystyle +(-2K-2M)[(e^{-2x}-2xe^{-2x})(sin2x)+(2xe^{-2x}cos2x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99fbec2b3dca4a5ddc2feb01cb741614a3bcdca8)
![{\displaystyle \displaystyle y(0)=1=c_{1}(1)+(1)[-{\frac {1}{4}}(1)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/525d7cfcaddc520cdb09cd4e1093a60d89942609)
![{\displaystyle \displaystyle y'(0)=-1.5=-2(-{\frac {5}{4}})(1)+c_{2}(1)-2(1)[-{\frac {1}{4}}(1)]+(1)(-1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b06d5447466e07865652a1a622cd98467d9841c4)
