0
$\begingroup$

how could I use Laplace Transform to solve the following differential equation:

$$y''+2y'+y=0; \;\;\;\;\;y'(0)=2\;\;\;\;\;\mbox{and}\;\;\;\;\;\boldsymbol{y(1)=2}.$$

The solution may involve the Heaviside step function, which is very similar to the Indicator function.

My problem is related to the initial value $y(1) = 2$. I had never seen a Laplace Transform problem with an initial value given at a point different from $0$.

I appreciate your help.

$\endgroup$
1
$\begingroup$

We are given:

$$y''+2y'+y=0; \;\;\;\;\;y'(0)=2\;\;\;\;\;\mbox{and}\;\;\;\;\;y(1)=2$$

I am going to start you off, but you have to show some work and fill in the details.

Taking the Laplace transform yields:

$$\mathscr{L}(y''+2y'+y) = (s^2 y(s) -s y(0) -y'(0)) +2 (s y(s) - y(0)) + y(s) = 0$$

Substituting the ICs (let $y(0) = a$) yields:

$$(s^2 y(s) -s a -2) +2 (s y(s) - a) + y(s) = 0$$

Solving for $y(s)$ yields:

$$y(s) = \dfrac{as +2a+2}{(s+1)^2}$$

Now, find the inverse Laplace transform:

$$y(t) = e^{-t}(at + a + 2t)$$

Next, substitute in $y(1) = 2$ and solve for $a$ yielding:

$$y(t) = e^{-t}(e t + t + e - 1)$$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.