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.


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:

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


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