Developing the second homogeneous solution for the case of double real root as a limiting case of distinct roots.
Consider two distinct roots of the form:
λ
1
=
x
{\displaystyle \lambda _{1}=x\!}
λ
2
=
x
+
ϵ
{\displaystyle \lambda _{2}=x+\epsilon \!}
(where
ϵ
{\displaystyle \epsilon \!}
Find the homogeneous L2-ODE-CC having the above distinct roots.
(
λ
−
λ
1
)
(
λ
−
(
λ
1
)
)
=
0
{\displaystyle (\lambda -\lambda _{1})(\lambda -(\lambda _{1}))=0\!}
(
λ
−
x
)
(
λ
−
(
x
+
ϵ
)
)
=
0
{\displaystyle (\lambda -x)(\lambda -(x+\epsilon ))=0\!}
λ
2
−
λ
x
−
λ
ϵ
−
λ
x
+
x
2
+
x
ϵ
=
0
{\displaystyle \lambda ^{2}-\lambda x-\lambda \epsilon -\lambda x+x^{2}+x\epsilon =0\!}
λ
2
−
λ
(
2
x
+
ϵ
)
+
x
(
x
+
ϵ
)
=
0
{\displaystyle \lambda ^{2}-\lambda (2x+\epsilon )+x(x+\epsilon )=0\!}
∴
y
″
−
y
′
(
2
λ
+
ϵ
)
+
y
λ
(
λ
+
ϵ
)
=
0
{\displaystyle \therefore y''-y'(2\lambda +\epsilon )+y\lambda (\lambda +\epsilon )=0\!}
Show that
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle {\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
Let's find the corresponding derivatives:
y
=
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
{\displaystyle y={\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}\!}
y
′
=
(
λ
+
ϵ
)
e
λ
+
ϵ
x
−
λ
e
λ
x
ϵ
{\displaystyle y'={\frac {(\lambda +\epsilon )e^{\lambda +\epsilon x}-\lambda e^{\lambda x}}{\epsilon }}\!}
y
″
=
(
λ
+
ϵ
)
2
e
(
λ
+
ϵ
)
x
−
λ
2
e
λ
x
ϵ
{\displaystyle y''={\frac {(\lambda +\epsilon )^{2}e^{(\lambda +\epsilon )x}-\lambda ^{2}e^{\lambda x}}{\epsilon }}\!}
If we now take these three equations and plug them into the homogeneous L2-ODE-CC (1), we get:
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 e^{(\lambda +\epsilon )x}(\lambda ^{2}+2\lambda \epsilon +\epsilon ^{2}-2\lambda ^{2}-2\lambda \epsilon -\lambda \epsilon -\epsilon ^{2}+\lambda ^{2}+\epsilon \lambda )+e^{\lambda x}(-\lambda ^{2}+2\lambda ^{2}+\lambda \epsilon -\lambda ^{2}-\lambda \epsilon )=0\!}
∴
0
≡
0.
{\displaystyle \therefore 0\equiv 0.\!}
Since the left and right hand sides of the equation are zero, the solution is in fact a homogeneous equation.
Find the limit of the homogeneous solution in (2) as epsilon approaches zero (think l'Hopital's Rule).
Using l'Hopital's Rule,
lim
ϵ
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
0
0
{\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}={\frac {0}{0}}\!}
(this is an indeterminate form).
L'Hopital's Rule states that we can divide this function into two functions,
f
(
ϵ
)
{\displaystyle f(\epsilon )\!}
g
(
ϵ
)
{\displaystyle g(\epsilon )\!}
f
′
(
ϵ
)
g
′
(
ϵ
)
{\displaystyle {\frac {f'(\epsilon )}{g'(\epsilon )}}\!}
lim
ϵ
→
0
f
′
(
ϵ
)
g
′
(
ϵ
)
=
lim
ϵ
→
0
x
e
x
(
λ
+
ϵ
)
1
{\displaystyle \lim _{\epsilon \rightarrow 0}{\frac {f'(\epsilon )}{g'(\epsilon )}}=\lim _{\epsilon \rightarrow 0}{\frac {xe^{x(\lambda +\epsilon )}}{1}}\!}
=
x
e
(
λ
+
0
)
x
1
{\displaystyle ={\frac {xe^{(\lambda +0)x}}{1}}\!}
=
x
e
x
λ
.
{\displaystyle =xe^{x\lambda }.\!}
Take the derivative of
e
λ
x
{\displaystyle e^{\lambda x}}
Taking the derivative with respect to lambda, we find that:
d
(
e
λ
x
)
d
λ
=
x
e
λ
x
{\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=xe^{\lambda x}}
It is important to remember that we must hold
x
{\displaystyle x\!}
Compare the results in parts (3) and (4), and relate to the result by using variation of parameters
Taking a closer look at Parts 3 and 4 of this problem, we discover that they're in fact equal:
d
(
e
λ
x
)
d
λ
=
lim
e
→
0
e
(
λ
+
ϵ
)
x
−
e
λ
x
ϵ
=
x
e
λ
x
{\displaystyle {\frac {d(e^{\lambda x})}{d\lambda }}=\lim _{e\rightarrow 0}{\frac {e^{(\lambda +\epsilon )x}-e^{\lambda x}}{\epsilon }}=xe^{\lambda x}\!}
Numerical experiment: Compute (2) setting lambda equal to 5 and epsilon equal to 0.001 </math>, and compare to the value obtained from the exact second homogeneous solution.
After performing these calculations, from (2) we get 148.478.
And from the exact second homogeneous solution, we get 200.05.
