Do Not Us This Page.
Go to User:Egm6321.f09.Team2/HW3 for latest information
Homework Assignment #3 - due Wednesday, 10/7, 21:00 UTC
Problem 1
edit
Find
(
m
,
n
)
{\displaystyle (m,n)}
p.13-1 ) is exact. A first integral is
Φ
(
x
,
y
,
p
)
=
x
p
+
(
2
x
3
2
−
1
)
y
+
k
1
=
k
2
{\displaystyle \Phi (x,y,p)=xp+(2x^{\frac {3}{2}}-1)y+k_{1}=k_{2}}
k
1
,
k
2
{\displaystyle k_{1},k_{2}}
Problem Statement : Given a L2_ODE_VC
x
y
″
+
2
x
y
′
+
3
y
=
0
{\displaystyle {\sqrt {x}}y''+2xy'+3y=0}
Find (m,n) from the integrating factor (xm ,yn ) that makes the equation exact.
A first integral is
ϕ
(
x
,
y
,
p
)
=
x
p
+
(
2
x
3
/
2
)
y
+
k
1
=
k
2
{\displaystyle \phi (x,y,p)=xp+{\big (}2x^{3/2}{\big )}y+k_{1}=k_{2}}
ϕ
p
=
x
{\displaystyle \phi _{p}=x}
ϕ
x
=
p
+
3
x
1
/
2
{\displaystyle \phi _{x}=p+3x^{1/2}}
ϕ
y
=
2
x
3
/
2
−
1
{\displaystyle \phi _{y}=2x^{3/2}-1}
ϕ
(
x
,
y
,
p
)
=
h
(
x
,
y
)
+
∫
f
(
x
,
y
,
p
)
d
p
,
f
=
ϕ
p
{\displaystyle \phi (x,y,p)=h(x,y)+\int _{}f(x,y,p)dp,f=\phi _{p}}
ϕ
(
x
,
y
,
p
)
=
h
(
x
,
y
)
+
∫
x
d
p
=
h
(
x
,
y
)
+
x
p
{\displaystyle \phi (x,y,p)=h(x,y)+\int _{}xdp={h(x,y)+xp}}
g
(
x
,
y
,
p
)
=
ϕ
x
+
ϕ
y
p
=
h
x
+
ϕ
x
+
(
h
y
+
ϕ
y
)
p
{\displaystyle g(x,y,p)=\phi _{x}+\phi _{y}p=h_{x}+\phi _{x}+(h_{y}+\phi _{y})p}
g
(
x
,
y
,
p
)
=
p
+
3
x
1
/
2
+
(
2
x
3
/
2
−
1
)
p
=
h
x
+
x
+
(
h
y
+
0
)
p
{\displaystyle g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_{x}+x+(h_{y}+0)p}
h
y
=
2
x
3
/
2
{\displaystyle h_{y}=2x^{3/2}}
h
x
=
3
x
1
/
2
−
x
{\displaystyle h_{x}=3x^{1/2}-x}
y
n
=
h
y
⇒
l
n
y
n
=
l
n
(
2
x
3
/
2
)
{\displaystyle y^{n}=h_{y}\Rightarrow lny^{n}=ln(2x^{3/2})}
n
=
l
n
(
2
x
3
/
2
)
{\displaystyle n=ln(2x^{3/2})}
x
m
=
h
x
⇒
l
n
x
m
=
l
n
(
3
x
1
/
2
−
x
)
{\displaystyle x^{m}=h_{x}\Rightarrow lnx^{m}=ln(3x^{1/2}-x)}
m
=
l
n
(
3
x
1
/
2
−
x
)
{\displaystyle m=ln(3x^{1/2}-x)}
Problem 2
edit
Solve eqn. 2 on (p.13-1 ) for
y
(
x
)
{\displaystyle y(x)}
Problem Statement: Given a first integral
ϕ
{\displaystyle \phi }
y
(
x
)
{\displaystyle y(x)}
ϕ
(
x
,
y
,
p
)
=
x
p
+
(
2
x
3
2
−
1
)
y
+
k
1
=
k
2
{\displaystyle \phi (x,y,p)=xp+(2x^{3 \over 2}-1)y+k_{1}=k_{2}}
where k1 and k2 are const, and
p
=
y
′
{\displaystyle p=y'}
Eq. (1) is in the form
M
(
x
,
y
)
+
N
(
x
,
y
)
y
′
{\displaystyle M(x,y)+N(x,y)y'}
M
(
x
,
y
)
=
(
2
x
3
2
−
1
)
y
{\displaystyle M(x,y)=(2x^{3 \over 2}-1)y}
N
(
x
,
y
)
=
x
{\displaystyle N(x,y)=x}
so it satisfies the 1st condition of exactness.
Check if
M
y
=
N
x
{\displaystyle M_{y}=N_{x}}
nd condition of exactness
M
y
=
2
x
3
2
−
1
{\displaystyle M_{y}=2x^{3 \over 2}-1}
N
x
=
1
{\displaystyle N_{x}=1}
M
y
≠
N
x
{\displaystyle M_{y}\neq N_{x}}
nd condition of exactness.
We must apply the integrating factor method for a L1_ODE_VC.
x
p
+
(
2
x
3
2
−
1
)
y
=
k
2
−
k
1
{\displaystyle xp+(2x^{3 \over 2}-1)y=k_{2}-k_{1}}
y
′
+
a
0
(
x
)
y
=
b
(
x
)
{\displaystyle y'+a_{0}(x)y=b(x)}
a
o
(
x
)
=
1
x
(
2
x
3
2
−
1
)
=
2
x
−
1
x
{\displaystyle a_{o}(x)={1 \over x}(2x^{3 \over 2}-1)=2{\sqrt {x}}-{1 \over x}}
b
(
x
)
=
k
2
−
k
1
x
{\displaystyle b(x)={k_{2}-k_{1} \over x}}
From our solution of a general non-homogeneous L1_ODE_VC p.8-1
h
(
x
)
=
e
x
p
∫
x
a
0
(
s
)
d
s
{\displaystyle h(x)=exp\int _{}^{x}a_{0}(s)ds}
h
(
x
)
=
e
x
p
∫
x
2
x
−
1
x
=
e
x
p
(
4
3
x
3
2
−
l
n
|
x
|
)
(
k
2
−
k
1
x
)
=
e
x
p
4
3
x
3
2
−
x
{\displaystyle h(x)=exp\int _{}^{x}2{\sqrt {x}}-{1 \over x}=exp{\bigg (}{4 \over 3}x^{3 \over 2}-ln\left|x\right|{\bigg )}{\bigg (}{{k_{2}-k_{1}} \over x}{\bigg )}=exp{4 \over 3}x^{3 \over 2}-x}
From p.8-2 Eq. (4)
y
(
x
)
=
1
h
(
x
)
∫
x
h
(
s
)
b
(
s
)
d
s
{\displaystyle y(x)={1 \over h(x)}\int _{}^{x}h(s)b(s)ds}
Use the product rule of integration
∫
a
b
=
a
∫
b
−
∫
a
′
∫
b
{\displaystyle \int _{}ab=a\int _{}b-\int _{}a'\int _{}b}
y
(
x
)
=
1
h
(
x
)
[
h
(
x
)
∫
x
b
(
x
)
−
∫
x
h
′
(
x
)
∫
x
b
(
x
)
]
{\displaystyle y(x)={1 \over h(x)}{\bigg [}h(x)\int _{}^{x}b(x)-\int _{}^{x}h'(x)\int _{}^{x}b(x){\bigg ]}}
In our example
∫
x
h
′
(
x
)
=
h
(
x
)
{\displaystyle \int _{}^{x}h'(x)=h(x)}
y
(
x
)
=
1
h
(
x
)
[
h
(
x
)
∫
x
b
(
x
)
−
h
(
x
)
∫
x
b
(
x
)
]
{\displaystyle y(x)={1 \over h(x)}{\bigg [}h(x)\int _{}^{x}b(x)-h(x)\int _{}^{x}b(x){\bigg ]}}
y
(
x
)
=
0
{\displaystyle y(x)=0}
Problem 3
edit
From (p.13-1 ), find the mathematical structure of
Φ
{\displaystyle \Phi }
F
(
x
,
y
,
y
′
,
y
″
)
=
d
ϕ
d
x
=
ϕ
x
(
x
,
y
,
p
)
+
ϕ
y
(
x
,
y
,
p
)
p
+
ϕ
p
(
x
,
y
,
p
)
p
′
w
h
e
r
e
p
=
y
′
{\displaystyle F(x,y,y',y'')={d\phi \over dx}=\phi _{x}(x,y,p)+\phi _{y}(x,y,p)p+\phi _{p}(x,y,p)p'\ where\ p=y'}
ϕ
=
h
(
x
,
y
)
+
∫
ϕ
p
d
p
=
h
(
x
,
y
)
+
ϕ
p
p
{\displaystyle \phi =h(x,y)+\int _{}\phi _{p}dp=h(x,y)+\phi _{p}p}
ϕ
x
=
h
x
+
ϕ
p
x
p
{\displaystyle \phi _{x}=h_{x}+\phi _{px}p}
ϕ
y
=
h
y
+
ϕ
p
y
p
{\displaystyle \phi _{y}=h_{y}+\phi _{py}p}
g
=
ϕ
x
+
ϕ
y
p
=
h
x
+
ϕ
p
x
p
+
(
h
y
+
ϕ
p
y
p
)
p
{\displaystyle g=\phi _{x}+\phi _{y}p=h_{x}+\phi _{px}p+(h_{y}+\phi _{py}p)p}
h
y
=
ϕ
y
−
ϕ
p
x
{\displaystyle h_{y}=\phi _{y}-\phi _{px}}
Take the integral of
h
y
{\displaystyle h_{y}}
h
(
x
,
y
)
=
(
ϕ
y
−
ϕ
p
x
)
y
+
k
1
{\displaystyle h(x,y)=(\phi _{y}-\phi _{px})y+k_{1}}
Substitute back into the equation for
ϕ
{\displaystyle \phi }
ϕ
=
(
ϕ
y
−
ϕ
p
x
)
y
+
k
1
+
ϕ
p
p
{\displaystyle \phi =(\phi _{y}-\phi _{px})y+k_{1}+\phi _{p}p}
Rearrange the terms to obtain
ϕ
(
x
,
y
,
p
)
=
ϕ
p
p
+
(
ϕ
y
−
ϕ
p
x
)
y
+
k
{\displaystyle \phi (x,y,p)=\phi _{p}p+(\phi _{y}-\phi _{px})y+k}
P
(
x
)
=
ϕ
p
{\displaystyle P(x)=\phi _{p}}
T
(
x
)
=
(
ϕ
y
−
ϕ
p
x
)
y
{\displaystyle T(x)=(\phi _{y}-\phi _{px})y}
k
=
k
1
{\displaystyle k=k_{1}}
ϕ
(
x
,
y
,
p
)
=
P
(
x
)
p
+
T
(
x
)
y
+
k
{\displaystyle \phi (x,y,p)=P(x)p+T(x)y+k}
Problem 4
edit
From (p.13-3 ), for the case
n
=
1
{\displaystyle n=1}
F
(
x
,
y
,
y
′
)
=
0
=
d
Φ
d
x
(
x
,
y
)
{\displaystyle F(x,y,y')=0={\frac {d\Phi }{dx}}(x,y)}
f
0
−
d
f
1
d
x
=
0
⇔
Φ
x
y
=
Φ
y
x
{\displaystyle f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \Phi _{xy}=\Phi _{yx}}
Hint: Use
f
1
=
Φ
y
{\displaystyle f_{1}=\Phi _{y}}
f
0
{\displaystyle f_{0}}
Φ
{\displaystyle \Phi }
f
1
{\displaystyle f_{1}}
Φ
{\displaystyle \Phi }
f
1
=
Φ
y
{\displaystyle f_{1}=\Phi _{y}}
f
0
−
d
f
1
d
x
=
0
⇔
Φ
x
y
=
Φ
y
x
{\displaystyle f_{0}-{\frac {df_{1}}{dx}}=0\Leftrightarrow \Phi _{xy}=\Phi _{yx}}
Problem Statement: Given a N1_ODE, for the case n=1
F
(
x
,
y
,
y
′
)
=
0
⇔
d
ϕ
d
x
(
x
,
y
)
{\displaystyle F(x,y,y')=0\Leftrightarrow {d\phi \over dx}(x,y)}
Show that
f
0
−
d
f
1
d
f
x
=
0
⇔
ϕ
x
y
=
ϕ
y
x
,
{\displaystyle f_{0}-{df_{1} \over df_{x}}=0\Leftrightarrow \phi _{xy}=\phi _{yx},}
f
1
=
ϕ
y
{\displaystyle f_{1}=\phi _{y}}
F
=
d
ϕ
d
x
(
x
,
y
(
0
)
,
.
.
.
.
y
(
n
−
1
)
=
ϕ
x
+
ϕ
y
6
(
0
)
y
(
1
)
{\displaystyle F={d\phi \over dx}(x,y^{(0)},....y^{(n-1)}=\phi _{x}+\phi _{y}6{(0)}y^{(1)}}
F
=
ϕ
x
+
ϕ
y
y
′
{\displaystyle F=\phi _{x}+\phi _{y}y'}
f
i
:=
∂
F
∂
y
i
{\displaystyle f_{i}:={\partial F \over \partial y^{i}}}
Find
f
0
{\displaystyle f_{0}}
ϕ
{\displaystyle \phi }
f
0
=
∂
F
∂
y
=
∂
(
ϕ
x
+
ϕ
y
y
′
)
∂
y
=
ϕ
x
y
{\displaystyle f_{0}={\partial F \over \partial y}={\partial (\phi _{x}+\phi _{y}y') \over \partial y}=\phi _{xy}}
f
0
=
ϕ
x
y
{\displaystyle f_{0}=\phi _{xy}}
Find
f
1
{\displaystyle f_{1}}
ϕ
y
{\displaystyle \phi _{y}}
f
1
=
∂
F
∂
y
′
=
∂
(
ϕ
x
+
ϕ
y
y
′
)
∂
y
′
=
ϕ
y
{\displaystyle f_{1}={\partial F \over \partial y'}={\partial (\phi _{x}+\phi _{y}y') \over \partial y'}=\phi _{y}}
f
1
=
ϕ
y
{\displaystyle f_{1}=\phi _{y}}
Show that
f
0
−
d
f
1
d
f
x
=
0
⇔
ϕ
x
y
=
ϕ
y
x
,
{\displaystyle f_{0}-{df_{1} \over df_{x}}=0\Leftrightarrow \phi _{xy}=\phi _{yx},}
ϕ
x
y
−
d
ϕ
y
d
f
x
=
ϕ
x
y
−
ϕ
y
x
=
0
{\displaystyle \phi _{xy}-{d\phi _{y} \over df_{x}}=\phi _{xy}-\phi _{yx}=0}
ϕ
x
y
=
ϕ
y
x
{\displaystyle \phi _{xy}=\phi _{yx}}
Problem 5
edit
Problem 6
edit
From (p.14-2 ), for the Legendre differential equation
F
=
(
1
−
x
2
)
y
″
−
2
x
y
′
+
n
(
n
+
1
)
y
=
0
{\displaystyle F=(1-x^{2})y''-2xy'+n(n+1)y=0}
p.10-3 ), Equations 4&5.p.14-1 ), Equation 5.
h
(
x
,
y
)
=
x
m
y
n
{\displaystyle h(x,y)=x^{m}y^{n}}
Problem 7
edit
Problem 8
edit
Problem 9
edit
Problem Statement: From (p.16-2 ), show that
y
x
x
x
=
e
−
3
t
(
y
t
t
t
−
3
y
t
t
+
2
y
t
)
{\displaystyle y_{xxx}=e^{-3t}\left(y_{ttt}-3y_{tt}+2y_{t}\right)}
y
x
x
x
x
=
e
−
4
t
(
y
t
t
t
t
−
6
y
t
t
t
+
11
y
t
t
−
6
y
t
)
{\displaystyle y_{xxxx}=e^{-4t}\left(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t}\right)}
y
x
x
x
=
(
d
/
d
x
)
[
d
/
d
x
(
d
/
d
x
)
y
]
=
(
d
t
/
d
x
)
(
d
/
d
t
)
(
d
t
/
d
x
)
(
d
/
d
t
)
(
d
t
/
d
x
)
(
d
/
d
t
)
y
{\displaystyle y_{xxx}=(d/dx){\bigg [}{d/dx \over (d/dx)y}{\bigg ]}=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y}
Replace
(
d
t
/
d
x
)
w
i
t
h
e
−
t
a
n
d
(
d
/
d
t
)
y
w
i
t
h
y
t
.
{\displaystyle (dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_{t}.}
y
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
d
/
d
t
)
(
(
e
−
t
)
y
t
)
{\displaystyle y_{xxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_{t})}
y
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
−
e
−
t
(
y
t
)
+
e
−
t
(
t
t
t
)
)
{\displaystyle y_{xxx}=(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_{t})+e^{-t}(t_{tt}))}
y
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
−
e
−
2
t
(
y
t
)
+
e
−
2
t
(
y
t
t
)
)
{\displaystyle y_{xxx}=(e^{-t})(d/dt)(-e^{-2t}(y_{t})+e^{-2t}(y_{tt}))}
y
x
x
x
=
(
e
−
t
)
(
2
e
−
2
t
(
y
t
)
−
e
−
2
t
(
y
t
t
)
−
2
e
−
2
t
(
y
t
t
)
+
e
−
2
t
(
y
t
t
t
)
)
{\displaystyle y_{xxx}=(e^{-t})(2e^{-2t}(y_{t})-e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))}
y
t
t
t
=
2
e
−
3
t
(
y
t
)
−
e
−
3
t
(
y
t
t
)
−
2
e
−
3
t
(
y
t
t
)
+
e
−
3
t
9
y
t
t
t
)
{\displaystyle y_{ttt}=2e-3t(y_{t})-e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}9y_{ttt})}
Factor out
e
−
3
t
{\displaystyle e^{-3t}}
y
x
x
x
=
(
e
−
3
t
)
(
y
t
t
t
−
3
y
t
t
+
2
y
t
)
{\displaystyle y_{xxx}=(e^{-3t})(y_{ttt}-3y_{tt}+2y_{t})}
y
x
x
x
x
=
(
d
/
d
x
)
(
d
/
d
x
)
(
d
/
d
x
)
(
d
/
d
x
)
y
{\displaystyle y_{xxxx}=(d/dx)(d/dx)(d/dx)(d/dx)y}
y
x
x
x
x
=
(
d
t
/
d
x
)
(
d
/
d
t
)
[
(
d
t
/
d
x
)
(
d
/
d
t
)
]
(
(
d
t
/
d
x
)
(
d
/
d
t
)
)
⟨
(
d
t
/
d
x
)
(
d
/
d
t
)
y
⟩
{\displaystyle y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))\langle (dt/dx)(d/dt)y\rangle }
Replace
(
d
t
/
d
x
)
w
i
t
h
e
−
t
a
n
d
(
d
/
d
t
)
y
w
i
t
h
y
t
.
{\displaystyle (dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_{t}.}
y
x
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
d
/
d
t
)
(
(
e
−
t
)
y
t
)
{\displaystyle y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_{t})}
y
x
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
−
e
−
t
(
y
t
)
+
e
−
t
(
y
t
t
)
)
{\displaystyle y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_{t})+e^{-t}(y_{tt}))}
y
x
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
d
/
d
t
)
(
e
−
2
t
(
y
t
)
+
e
−
2
t
(
y
t
t
)
)
{\displaystyle y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-2t}(y_{t})+e^{-2t}(y_{tt}))}
y
x
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
e
−
t
)
(
−
2
e
−
2
t
(
y
t
)
+
e
−
2
t
(
y
t
t
)
−
2
e
−
2
t
(
y
t
t
)
+
e
−
2
t
(
y
t
t
t
)
)
{\displaystyle y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(-2e^{-2t}(y_{t})+e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))}
y
x
x
x
x
=
(
e
−
t
)
(
d
/
d
t
)
(
−
2
e
−
3
t
(
y
t
)
+
e
−
3
t
(
y
t
t
)
−
2
e
−
3
t
(
y
t
t
)
+
e
−
3
t
(
y
t
t
t
)
)
{\displaystyle y_{xxxx}=(e^{-t})(d/dt)(-2e^{-3t}(y_{t})+e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}(y_{ttt}))}
y
x
x
x
x
=
(
e
−
t
)
(
6
e
−
3
t
(
y
t
)
−
2
e
−
3
t
(
y
t
t
)
−
3
e
−
3
t
(
y
t
t
)
+
e
−
3
t
(
y
t
t
t
)
+
6
e
−
3
t
(
y
t
t
)
−
2
e
−
3
t
(
y
t
t
t
)
−
3
e
−
3
t
(
y
t
t
t
)
+
e
−
3
t
(
y
t
t
t
t
)
)
{\displaystyle y_{xxxx}=(e^{-t})(6e^{-3t}(y_{t})-2e^{-3t}(y_{tt})-3e^{-3t}(y_{tt})+e^{-3t}(y_{ttt})+6e^{-3t}(y_{tt})-2e^{-3t}(y_{ttt})-3e^{-3t}(y_{ttt})+e^{-3t}(y_{tttt}))}
y
x
x
x
x
=
6
e
−
4
t
(
y
t
)
+
2
e
−
4
t
(
y
t
t
)
+
3
e
−
4
t
(
y
t
t
)
−
e
−
4
t
(
y
t
t
t
)
+
6
e
−
4
t
(
y
t
t
)
−
2
e
−
4
t
(
y
t
t
t
)
−
3
e
−
4
t
(
y
t
t
t
)
+
e
−
4
t
(
y
t
t
t
t
)
)
{\displaystyle y_{xxxx}=6e^{-4t}(y_{t})+2e^{-4t}(y_{tt})+3e^{-4t}(y_{tt})-e^{-4t}(y_{ttt})+6e^{-4t}(y_{tt})-2e^{-4t}(y_{ttt})-3e^{-4t}(y_{ttt})+e^{-4t}(y_{tttt}))}
Factor out
e
−
4
t
{\displaystyle e^{-4t}}
y
x
x
x
x
=
(
e
−
4
t
)
(
y
t
t
t
t
−
6
y
t
t
t
+
11
y
t
t
−
6
y
t
)
{\displaystyle y_{xxxx}=(e^{-4t})(y_{tttt}-6y_{ttt}+11y_{tt}-6y_{t})}
Problem 10
edit
Problem 11
edit
Problem Statement: From (p.17-4 ) obtain equation 2 from p.17-3
Z
(
x
)
=
c
u
1
2
exp
(
−
∫
x
a
1
(
s
)
d
s
)
{\displaystyle Z(x)={\frac {c}{u_{1}^{2}}}\exp \left(-\int ^{x}a_{1}(s)ds\right)}
Problem 12
edit
Problem Statement: From (p.18-1 ), develop reduction of order method using the following algebraic options
y
(
x
)
=
U
(
x
)
±
u
1
(
x
)
{\displaystyle y(x)=U(x)\pm u_{1}(x)}
y
(
x
)
=
U
(
x
)
u
1
(
x
)
{\displaystyle y(x)={\frac {U(x)}{u_{1}(x)}}}
y
(
x
)
=
u
1
(
x
)
U
(
x
)
{\displaystyle y(x)={\frac {u_{1}(x)}{U(x)}}}
Problem 13
edit
Problem Statement: From (p.18-1 ), Find
u
1
(
x
)
{\displaystyle u_{1}(x)}
u
2
(
x
)
{\displaystyle u_{2}(x)}
y
=
a
x
b
{\displaystyle y=ax^{b}}
y
=
e
r
x
{\displaystyle y=e^{rx}}
Compare the two solutions using boundary conditions
y
(
0
)
=
1
{\displaystyle y(0)=1}
y
(
1
)
=
2
{\displaystyle y(1)=2}
Contributing Team Members
edit
![{\displaystyle y(x)={1 \over h(x)}{\bigg [}h(x)\int _{}^{x}b(x)-\int _{}^{x}h'(x)\int _{}^{x}b(x){\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1fc9b6bfe26c1f496c8199f5185c20147ad72a9)
![{\displaystyle y(x)={1 \over h(x)}{\bigg [}h(x)\int _{}^{x}b(x)-h(x)\int _{}^{x}b(x){\bigg ]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e5c4089dc28b61daa8da4b8d2db5798137c66be)
![{\displaystyle y_{xxx}=(d/dx){\bigg [}{d/dx \over (d/dx)y}{\bigg ]}=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f574942ac5ee0c61b7a479e49a2ba95a5a2183d)
(d/dt))\langle (dt/dx)(d/dt)y\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a03077d4ff5cec982477e1504c00bc905f7f74b)
