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

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 1

It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:

$y + p y' + q y = \map R x$

where:

$p = -1$

$q = -6$

$\map R x = e^{-x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y - y' - 6 y = 0$

From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:

$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$

We have that:

$\map R x = e^{-x}$

and so from the Method of Undetermined Coefficients for the Exponential function:

$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$

where:

$K = 1$

$a = -1$

$p = -1$

$q = -6$

Hence:

\(\ds y_p\)

\(=\)

\(\ds \dfrac {e^{-x} } {1 + 1 - 6}\)

\(\ds \)

\(=\)

\(\ds -\dfrac {e^{-x} } 4\)

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

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

$\blacksquare$

Proof 2

It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:

$y + p y' + q y = \map R x$

where:

$p = -1$

$q = -6$

$\map R x = e^{-x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y - y' - 6 y = 0$

From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:

$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$

It remains to find a particular solution $y_p$ to $(1)$.

Expressing $y_g$ in the form:

$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$

we have:

\(\ds \map {y_1} x\)

\(=\)

\(\ds e^{3 x}\)

\(\ds \map {y_2} x\)

\(=\)

\(\ds e^{-2 x}\)

\(\ds \leadsto \ \ \)

\(\ds \map { {y_1}'} x\)

\(=\)

\(\ds 3 e^{3 x}\)

Derivative of Exponential Function

\(\ds \map { {y_2}'} x\)

\(=\)

\(\ds -2 e^{-2 x}\)

Derivative of Exponential Function

By the Method of Variation of Parameters, we have that:

$y_p = v_1 y_1 + v_2 y_2$

where:

\(\ds v_1\)

\(=\)

\(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)

\(\ds v_2\)

\(=\)

\(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)

where $\map W {y_1, y_2}$ is the Wronskian of $y_1$ and $y_2$.

We have that:

\(\ds \map W {y_1, y_2}\)

\(=\)

\(\ds y_1 {y_2}' - y_2 {y_1}'\)

Definition of Wronskian

\(\ds \)

\(=\)

\(\ds e^{3 x} \paren {-2 e^{-2 x} } - e^{-2 x} \paren {3 e^{3 x} }\)

\(\ds \)

\(=\)

\(\ds -2 e^x - 3 e^x\)

\(\ds \)

\(=\)

\(\ds -5 e^x\)

Hence:

\(\ds v_1\)

\(=\)

\(\ds \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x\)

\(\ds \)

\(=\)

\(\ds \int -\frac {e^{-2 x} \cdot e^{-x} } {-5 e^x} \rd x\)

\(\ds \)

\(=\)

\(\ds \int \frac {e^{-4 x} } 5 \rd x\)

\(\ds \)

\(=\)

\(\ds -\frac 1 {20} e^{-4 x}\)

Primitive of $e^{a x}$

\(\ds v_2\)

\(=\)

\(\ds \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x\)

\(\ds \)

\(=\)

\(\ds \int \frac {e^{3 x} \cdot e^{-x} } {-5 e^x} \rd x\)

\(\ds \)

\(=\)

\(\ds \int -\frac {e^x} 5 \rd x\)

\(\ds \)

\(=\)

\(\ds -\frac 1 5 e^x\)

Primitive of $e^{a x}$

It follows that:

\(\ds y_p\)

\(=\)

\(\ds -\frac 1 {20} e^{-4 x} e^{3 x} - \frac 1 5 e^x e^{-2 x}\)

\(\ds \)

\(=\)

\(\ds -\frac 1 {20} e^{-x} - \frac 1 5 e^{-x}\)

simplifying

\(\ds \)

\(=\)

\(\ds -\frac 1 4 e^{-x}\)

further simplifying

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

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

is the general solution to $(1)$.

$\blacksquare$

Proof 3

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.

$\blacksquare$