Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 3

Theorem

The second order ODE:

$(1): \quad y - y' - 6 y = e^{-x}$

has the general solution:

$y = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$

Proof

Taking Laplace transforms:

$\laptrans {y - y' - 6 y} = \laptrans {e^{-x} }$

We have:

\(\ds \laptrans {y - y' - 6 y}\)

\(=\)

\(\ds \laptrans {y} - \laptrans {y'} - 6 \laptrans y\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds s^2 \laptrans y - s \map y 0 - \map {y'} 0 - \paren {s \laptrans y - \map y 0} - 6 \laptrans y\)

\(\ds \)

\(=\)

\(\ds \paren {s^2 - s - 6} \laptrans y - \paren {s \map y 0 + \map {y'} 0 - \map y 0}\)

We also have:

\(\ds \laptrans {e^{-x} }\)

\(=\)

\(\ds \frac 1 {s - \paren {-1} }\)

Laplace Transform of Exponential

\(\ds \)

\(=\)

\(\ds \frac 1 {s + 1}\)

So:

$\paren {s^2 - s - 6} \laptrans y = s \map y 0 + \paren {\map {y'} 0 - \map y 0} + \dfrac 1 {s + 1}$

Giving:

\(\ds \laptrans y\)

\(=\)

\(\ds \map y 0 \paren {\frac s {s^2 - s - 6} } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {s^2 - s - 6} } + \frac 1 {\paren {s + 1} \paren {s^2 - s - 6} }\)

\(\ds \)

\(=\)

\(\ds \map y 0 \paren {\frac s {\paren {s - 3} \paren {s + 2} } } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {\paren {s - 3} \paren {s + 2} } } + \frac 1 {\paren {s + 1} \paren {s - 3} \paren {s + 2} }\)

factorising

\(\ds \)

\(=\)

\(\ds \frac {\map y 0} 5 \paren {\frac 2 {s + 2} + \frac 3 {s - 3} } + \frac {\map {y'} 0 - \map y 0} 5 \paren {\frac 1 {s - 3} - \frac 1 {s + 2} } + \frac 1 {20} \paren {\frac 1 {s - 3} + \frac 4 {s + 2} - \frac 5 {s + 1} }\)

partial fraction expansion

\(\ds \)

\(=\)

\(\ds \paren {\frac {2 \map y 0 - \map {y'} 0 + \map y 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {3 \map y 0 + \map {y'} 0 - \map y 0} 5 + \frac 1 {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }\)

\(\ds \)

\(=\)

\(\ds \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }\)

So:

\(\ds y\)

\(=\)

\(\ds \invlaptrans {\paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} } }\)

Definition of Inverse Laplace Transform

\(\ds \)

\(=\)

\(\ds \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \invlaptrans {\frac 1 {s + 2} } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \invlaptrans {\frac 1 {s - 3} } - \frac 1 4 \invlaptrans {\frac 1 {s + 1} }\)

Linear Combination of Laplace Transforms

\(\ds \)

\(=\)

\(\ds \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \invlaptrans {\laptrans {e^{-2 x} } } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \invlaptrans {\laptrans {e^{3 x} } } - \frac 1 4 \invlaptrans {\laptrans {e^{-x} } }\)

Laplace Transform of Exponential

\(\ds \)

\(=\)

\(\ds \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} e^{-2 x} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } e^{3 x} - \frac 1 4 e^{-x}\)

Definition of Inverse Laplace Transform

Setting:

$C_1 = \dfrac {8 \map y 0 + 4 \map {y'} 0 + 1} {20}$

$C_2 = \dfrac {3 \map y 0 - \map {y'} 0 + 1} 5$

gives the result.

$\blacksquare$