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

In summary: Sorry, I made a mistake in my previous post. Here's the corrected version:In summary, we have to solve the differential equation with initial conditions of $y'' - 5y' - 6y = e^{3x}$, $y(0) = 2$, and $y'(0) = 1$. By applying the Laplace transform, we get $s^2 Y(s) - (sy(0) + y'(0)) - 5s Y(s) + y(0) - 6Y(s) = 1/(s-3)$. This simplifies to $Y(s) = (2s^2 - 15s + 27)/(s-3)(
  • #1
351
87
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?
 
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  • #2
For starters, [tex]
\frac{1}{s-3} + 2s - 9 = \frac{1 + (2s-9)(s-3)}{s-3} = \frac{2s^2 - 15s + 28}{s-3}.[/tex]
 
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  • #3
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.
 
  • #4
May I suggest using Wolfram alpha or similar (Mathematica is great if you have access to it) to double check arithmetic?
 
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  • #5
Hall said:
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}##
 
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  • #6
Hall said:
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.
 
  • #7
Hall said:
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)##…
 
  • #8
chwala said:
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.
 
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1. How do I use Laplace Transform to solve a differential equation?

To use Laplace Transform, you first need to take the Laplace Transform of both sides of the differential equation. This will give you an algebraic equation in terms of the Laplace Transform of the function. Then, you can use algebraic methods to solve for the Laplace Transform of the function. Finally, you can take the inverse Laplace Transform to get the solution to the original differential equation.

2. What is the Laplace Transform of a derivative?

The Laplace Transform of a derivative is equal to the product of the Laplace Transform of the function and the initial value of the function. In other words, if the Laplace Transform of a function is F(s), then the Laplace Transform of its derivative is sF(s) - f(0).

3. How do I handle the e^{3x} term in the given differential equation?

To handle the e^{3x} term, you can use the property of Laplace Transform that states e^{at} transforms to 1/(s-a). In this case, the e^{3x} term will transform to 1/(s-3).

4. What do I do with the constants in the differential equation?

The constants in the differential equation, such as the -5 and -6 in the given equation, will also transform to constants in the algebraic equation. You can then solve for these constants using algebraic methods.

5. How do I know if I have the correct solution using Laplace Transform?

To check if you have the correct solution, you can take the Laplace Transform of your solution and see if it matches the algebraic equation you obtained by taking the Laplace Transform of the given differential equation. You can also plug in your solution to the original differential equation and see if it satisfies the equation.

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