Question

(a) Use appropriate algebra and Theorem to find the given inverse Laplace transform. (Write your answer as a function of t.)

L−1 {3s − 10/ s2 + 25}
(b) Use the Laplace transform to solve the given initial-value problem.
y' + 3y = e6t, y(0) = 2

Answer 1

(a) Expand the given expression as

`(3s-10)/(s^2+25)=3\cdot\frac s{s^2+25}-2\cdot\frac5{s^2+25}`

You should recognize the Laplace transform of sine and cosine:

`L[\cos(at)]=\frac s{s^2+a^2}`

`L[\sin(at)]=\frac a{s^2+a^2}`

So we have

`L^(-1)\left[(3s-10)/(s^2+25)\right]=3\cos(5t)-2\sin(5t)`

(b) Take the Laplace transform of both sides:

`y'(t)+3y(t)=e^(6t)\implies (sY(s)-y(0))+3Y(s)=\frac1{s-6}`

Solve for
`Y(s)`:

`(s+3)Y(s)-2=\frac1{s-6}\implies Y(s)=(2s-11)/((s-6)(s+3))`

Decompose the right side into partial fractions:

`(2s-11)/((s-6)(s+3))=(\theta_1)/(s-6)+(\theta_2)/(s+3)`

`2s-11=\theta_1(s+3)+\theta_2(s-6)`

`2s-11=(\theta_1+\theta_2)s+(3\theta_1-6\theta_2)`

`\begin{cases}\theta_1+\theta_2=2\\3\theta_1-6\theta_2=-11\end{cases}\implies\theta_1=\frac19,\theta_2=\frac{17}9`

So we have

`Y(s)=\frac19\cdot\frac1{s-6}+\frac{17}9\cdot\frac1{s+3}`

and taking the inverse transforms of both sides gives

`y(t)=\frac19e^(6t)+\frac{17}9e^(-3t)`