Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x/Proof 2
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Theorem
The second order ODE:
- $(1): \quad y'' + 4 y = 8 x^2 - 4 x$
has the general solution:
- $y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
Proof
Taking Laplace transforms:
- $\laptrans {y'' + 4 y} = \laptrans {8 x^2 - 4 x}$
We have:
\(\ds \laptrans {y'' + 4 y}\) | \(=\) | \(\ds \laptrans {y''} + 4 \laptrans y\) | Linear Combination of Laplace Transforms | |||||||||||
\(\ds \) | \(=\) | \(\ds s^2 \laptrans y - s \map y 0 - \map {y'} 0\) | Laplace Transform of Second Derivative |
We also have:
\(\ds \laptrans {8 x^2 - 4 x}\) | \(=\) | \(\ds 8 \laptrans {x^2} - 4 \laptrans x\) | Linear Combination of Laplace Transforms | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {8 \cdot 2!} {s^3} - \frac {4} {s^2}\) | Laplace Transform of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {16 - 4 s} {s^3}\) |
So:
- $\paren {s^2 + 4} \laptrans y = s \map y 0 + \map {y'} 0 + \dfrac {16 - 4 s} {s^3}$
Giving:
\(\ds \laptrans y\) | \(=\) | \(\ds \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac {16 - 4 s} {s^3 \paren {s^2 + 4} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac 4 {s^3} + \frac {s + 1} {s^2 + 4} - \frac 1 {s^2} - \frac 1 s\) | partial fraction expansion | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s\) |
So:
\(\ds y\) | \(=\) | \(\ds \invlaptrans {\paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s}\) | Definition of Inverse Laplace Transform | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map y 0 + 1} \invlaptrans {\frac s {s^2 + 2^2} } + \frac {\paren {\map {y'} 0 + 1} } 2 \invlaptrans {\frac 2 {s^2 + 2^2} } + 2 \invlaptrans {\frac {2!} {s^3} } - \invlaptrans {\frac {1!} {s^2} } - \invlaptrans {\frac {0!} s}\) | Linear Combination of Laplace Transforms | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map y 0 + 1} \invlaptrans {\laptrans {\cos 2 x} } + \frac {\paren {\map {y'} 0 + 1} } 2 \invlaptrans {\laptrans {\sin 2 x} } + 2 \invlaptrans {\laptrans {x^2} } - \invlaptrans {\laptrans x} - \invlaptrans {\laptrans 1}\) | Laplace Transform of Cosine, Laplace Transform of Sine, Laplace Transform of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\map y 0 + 1} \cos 2 x + \paren {\frac {\paren {\map {y'} 0 + 1} } 2 } \sin 2 x - 1 - x + 2 x^2\) | Definition of Inverse Laplace Transform |
Setting $C_1 = \dfrac {\paren {\map {y'} 0 + 1} } 2$ and $C_2 = \map y 0 + 1$ gives the result.
$\blacksquare$