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.

$\blacksquare$