Solving ##y'' - 5 y' - 6y = e^{3x}## using Laplace Transform

[Hall]

Homework Statement

Nil

Relevant Equations

Nil

We have to solve
$$
\begin{align*}
y'' - 5y' - 6y = e^{3x} \\
y(0) = 2,~~ y'(0) = 1 \\
\end{align*}
$$
Applying Laplace Transform the equation
$$
\begin{align*}
L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\
s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\
Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\
Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\
\textrm{On Partial Fraction Decomposition}\\
\frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\
A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\
\textrm{putting s =6} \\
21C = 9 \implies C = \frac{3}{7} \\
\textrm{similarly,} \\
s = -1 ~\text{gives}~ B = 11/7 \\
s = 3 ~ \text{gives} ~ A = 0\\
\textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\
y_p = 11/7 e^{-x} + 3/7 e^{6x}
\end{align*}
$$

But this doesn't match with the answer given in the book. Where is my mistake?

 

[pasmith]

For starters, $$\frac{1}{s-3} + 2s - 9 = \frac{1 + (2s-9)(s-3)}{s-3} = \frac{2s^2 - 15s + 28}{s-3}.$$

 

[FactChecker]

At the end of your third line, I think I get a -1 instead of a -9.
In general, it would be good for you to double-check your arithmetic.

 

[Orodruin]

May I suggest using Wolfram alpha or similar (Mathematica is great if you have access to it) to double check arithmetic?

 

[chwala]

Homework Statement:: Nil
Relevant Equations:: Nil

We have to solve
$$
\begin{align*}
y'' - 5y' - 6y = e^{3x} \\
y(0) = 2,~~ y'(0) = 1 \\
\end{align*}
$$
Applying Laplace Transform the equation
$$
\begin{align*}
L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\
s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\
Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\
Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\
\textrm{On Partial Fraction Decomposition}\\
\frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\
A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\
\textrm{putting s =6} \\
21C = 9 \implies C = \frac{3}{7} \\
\textrm{similarly,} \\
s = -1 ~\text{gives}~ B = 11/7 \\
s = 3 ~ \text{gives} ~ A = 0\\
\textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\
y_p = 11/7 e^{-x} + 3/7 e^{6x}
\end{align*}
$$

But this doesn't match with the answer given in the book. Where is my mistake?

I find the method of characteristic to be straightforward ...i.e

Let

$m^2-5m-6=0$

$m_1=-1$ $m_2=6$ therefore,

$y = Ae^{-x} + Be^{6x}$ on applying boundary condition; $y(0)=2$ we shall have

$2=A+B$
and on applying the second boundary condition; $y^{'}(0)=1$ we shall have

$1=-A+6B$

solving the simultaneous equation;

$2=A+B$
$1=-A+6B$

yields;

$A=\dfrac{11}{7}$ and $B=\dfrac{3}{7}$ therefore our complementary solution is;

$y_c(x) =\dfrac{11}{7} e^{-x} + \dfrac{3}{7} e^{6x}$ now on the inhomogenous part;

Let $y_p(x)= Ce^{3x}$

$y^{'}_p(x)= 3Ce^{3x}$

$y^{''}_p(x)= 9Ce^{3x}$ therefore;

$9Ce^{3x}-15Ce^{3x}-6Ce^{3x}= e^{3x}$

$-12Ce^{3x}=e^{3x}$

$C=-\dfrac{1}{12}$

Therefore $y(x)= y_c(x) + y_p(x)= \dfrac{11}{7} e^{-x} + \dfrac{3}{7} e^{6x}-\dfrac{1}{12}e^{3x}$

 

[chwala]

Homework Statement:: Nil
Relevant Equations:: Nil

We have to solve
$$
\begin{align*}
y'' - 5y' - 6y = e^{3x} \\
y(0) = 2,~~ y'(0) = 1 \\
\end{align*}
$$
Applying Laplace Transform the equation
$$
\begin{align*}
L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\
s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\
Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\
Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\
\textrm{On Partial Fraction Decomposition}\\
\frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\
A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\
\textrm{putting s =6} \\
21C = 9 \implies C = \frac{3}{7} \\
\textrm{similarly,} \\
s = -1 ~\text{gives}~ B = 11/7 \\
s = 3 ~ \text{gives} ~ A = 0\\
\textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\
y_p = 11/7 e^{-x} + 3/7 e^{6x}
\end{align*}
$$

But this doesn't match with the answer given in the book. Where is my mistake?

Where is your solution for the inhomogenous part?...your answer does not look complete to me! just check if my solution is correct. Cheers.

 

[Orodruin]

Homework Statement:: Nil
Relevant Equations:: Nil

We have to solve
$$
\begin{align*}
y'' - 5y' - 6y = e^{3x} \\
y(0) = 2,~~ y'(0) = 1 \\
\end{align*}
$$
Applying Laplace Transform the equation
$$
\begin{align*}
L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\
s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\
Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\
Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\
\textrm{On Partial Fraction Decomposition}\\
\frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\
A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\
\textrm{putting s =6} \\
21C = 9 \implies C = \frac{3}{7} \\
\textrm{similarly,} \\
s = -1 ~\text{gives}~ B = 11/7 \\
s = 3 ~ \text{gives} ~ A = 0\\
\textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\
y_p = 11/7 e^{-x} + 3/7 e^{6x}
\end{align*}
$$

But this doesn't match with the answer given in the book. Where is my mistake?

Taking the Laplace transform of $5y’$ you have transformed this to $5sY(s) - y(0)$…

 

[Orodruin]

Where is your solution for the inhomogenous part?...your answer does not look complete to me! just check if my solution is correct. Cheers.

The Laplace transform gets both the homogeneous and inhomogeneous parts since it solves the differential equation including the initial conditions.

Yes, you can use the characteristic equation, but the OP specifically asked about the Laplace transform method.