Question

write the solution of ivp in terms of a convolution integral :
y"+y'+5/4y= 1-u(t-pi) with initial conditions y(0)=1 and y'(0)=-1

Answer 1

It appears my response will be late here.

I am wondering what you have tried on this problem so far! But I will settle to provide a general response.

My thoughts so far:
We should first take the Laplace transform of both sides. Then solve for the Laplace transform of y in terms of s, the other variable. After we have the Laplace of y in terms of s, we should find two functions that multiply together to create the right side. Thus we can use the theorem:
\( \mathcal{L} \left[ f(t) * g(t) \right] (s) = \mathcal{L} \left[ f(t) \right] \cdot \mathcal{L} \left[ g(t) \right] = F(s) \cdot G(s) \)
where \(*\) denotes convolution of the two functions and \(\cdot\) represents regular multiplication. If we work backwards, we can obtain the Laplace of a convolution, which means when we apply the inverse Laplace transform to solve for y in terms of t, we obtain a convolution of two functions in the end. After all that, we just rewrite the convolution as a convolution integral expression using the definition:
\(\displaystyle f(t) * g(t) = \int_{0}^{t} f(t) \cdot g(t - \tau) d \tau \)

Good luck! And please bump the question, reply, and tag someone to help when you return if you require further assistance! Thanks :)