Question

Use the Laplace transform to solve the following initial value problem:
\(2y"+4y'+17y=3\cos(2t)\)
\(y(0)=y'(0)=0\)
a)take Laplace transform of both sides of the given differntial equation to create corresponding algebraic equation and then solve for \(L\left\{y(t)\right\}\) b) Express the solution \(y(t)\) in terms of a convolution integral

Answer 1

Step 1
Given differntial equation is,
\(2y"+4y'+17y=3\cos(2t)\)
\(y(0)=y'(0)=0\)
\(L\left[y"\right]=s^2L\left\{y(t)\right\}-sy(0)-y'(0)\)
\(L\left[y'\right]=sL\left\{y(t)\right\}-y(0)\)
\(L\left[\cos(at)\right]=\frac{s}{s^2+a^2}\)
Taking Laplace transform of equation (1),
\(2L\left[y"\right]+4L\left[y'\right]+17L\left[y\right]=3L\left[\cos(2t)\right]\)
\(s^2L\left\{y(t)\right\}-sy(0)-y'(0)+4\left[sL\left\{y(t)\right\}-y(0)\right]+17L\left\{y(t)\right\}=3L\left\{\cos(2t)\right\}\)
\(s^2Ly(t)+4sLy(t)+17Ly(t)=3x\frac{s}{s^2+4}\)
\((s^2+4s+17)L\left\{y(t)\right\}=\frac{3s}{s^2+4}\)
\(a) \therefore L\left\{y(t)\right\}=\frac{3s}{s^2+4} \times \frac{1}{s^2+4s+17}\)
\(b) y(t)=\int_0^t \left[3\cos(3w)\right] \times \left[e^{-2t} \cdot \sin \sqrt{13}w\right] dw\)
This is required Laplace transform.