Using Stepfunction to solve IVP

[Susanne217]

Homework Statement

I am given the task for finding the laplace transformation given

$$y'' + 4y = 8t^2$$ if 0<t<5 and 0 if t>5 y(1) = 1+cos(2) $$y'(1)= 4-2sin(2)$$

Homework Equations

The Attempt at a Solution

I know that the above problem can be written $$y''+4y=8 \cdot u(t)$$

where u(t) is the step function.

But what is my next step?

$$\mathcal{L}(y''+4y) = \mathcal{L}(8t^2)$$

which equals

$$\mathcal{L}(y'') + \mathcal{L}(4y) = 8\mathcal{L}(t^2)$$

$$\mathcal{L}(y'') = s \cdot \mathcal{L}(y') - y'(1)$$

$$\mathcal{L}(y') = s \cdot \mathcal{L}(y) - y(1)$$

not sure what to do about the left side by the rightside that is $$Y(s^2+ 4) = \frac{16}{s^3} + s(1+cos(2)) + (4-2sin(2))$$

then $$Q(s)= \frac{1}{s^2+4}$$

then $$Y(s) = s(1+cos(2)) + 4-2sin(2)) \cdot \frac{1}{s^2+4} + \frac{16}{s^3} \cdot \frac{1}{s^2+4}$$ Is this true or am I totally fubar?

 

[mighty2000]

Your approach is correct so far. The next step would be to use partial fraction decomposition to simplify the expression for Y(s) and then take the inverse Laplace transform to get the solution in the time domain. You can also use the initial conditions y(1) and y'(1) to solve for the constants in the partial fraction decomposition. Keep in mind that the Laplace transform of the step function u(t) is 1/s, so you will need to use this in your solution. Overall, your approach seems to be on the right track.