Transforming a Second Order Differential Equation Using Laplace Transforms

[jaejoon89]

Homework Statement

Solve by Laplace transforms the following
y'' + y = t when 0</=t<1, and = 1 if t>/=1

Homework Equations

L{y''} + L{y} = L{f(t)}

The Attempt at a Solution

By Laplace transforms I get
L{f(t)} = (1 - e^-s) / s^2
and
Y(s) = (1-e^-2 + s^2) / s^2 (s^2 +1)

But I cannot simplify Y(s) in order to get y = L^-1{Y(s)}!

 

[Unco]

Hi Jaejoon,

You didn't tell us what the initial conditions are, but from your work I'll assume they were that y(0)=0 and y'(0)=1.

To find the inverse Laplace transform of Y(s) = \frac{1-e^{-s} + s^2}{s^2(s^2 +1)}, first split it up to

\frac{1}{s^2} - \frac{e^{-s}}{s^2(s^2 +1)}.

To transform the second term, use the formula e^{-cs}F(s) \mapsto u(t-c)f(t-c), where u is the Heaviside function and f is the inverse Laplace of F. Note that you will need to split F(s) into partial fractions to transform using tables.

 

[jaejoon89]

Thanks.