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I was trying to solve a differential equation with the Laplace transform.

The following equation: xy''+(2x+3)y'+(x+3)y=3e^-x, y(0)=0.

L[xy'']= -d/dp(p^2Y-py(0)) L[3e^-x]=3/s+1 This is what I have solved.

The issue I am having is finding L[(2x+3)y'] and L[(x+3)y]

Anonymous
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1 Answers1

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For this question, you will need the following properties:

  1. $L[f'] = pL[f] - f(0)$
  2. $L[x^nf(x)] = (-1)^n F(p)$ where $F(p) = L[f]$

Using these, and a basic transformations you can get:

  1. $L[xy''] = (-1)^1 L'[y''] = \frac{-d}{dp}(p^2L[y]) = (-1)(2pY + p^2Y')$

  2. $L[xy'] = -(Y+pY')$

  3. $L[xy] = -Y'$

  4. $L[y'] = pL[y] = pY$

  5. $L[e^{-ax}] = \frac{1}{p+a}$

Breaking up the equation as $xy'' + 2xy' + 3y' - xy + 3y = 3e^{-x}$, we get the following equation after taking Laplace transformation on both sides:

$Y'(-p^2-2p-1) +(p+1)Y = 3/(p+1)$

This is a typical example of how Laplace transformation helps in reducing a second order ODE to first order. Solve it with the integrating factor $e^{\mathop{\int {\frac{-1}{p+1} dp}}} $ to get

$Y = L[y] = \frac{6}{(p+1)^2}$ $\implies y = L^{-1}[\frac{6}{(p+1)^2}]$

Now you need the shift rule for inverse Laplace transform:

  1. $L^{-1}[F(p+a)] = e^{-ax}L^{-1}[F(p)]$, and

  2. $L^{-1}[\frac{1}{p^{n+1}}] = \frac{x^{n}}{n!}$

$\implies y = e^{-x}L^{-1}[\frac{6}{(p)^2}] = xe^{-x}$

Do tell if I have made any calc errors.

Ashman Mehra
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