Question

4> Solve by using Laplace transform: y'+5y'+4y=0; y(0)=3 y'(o)=o

Answer 1

`y=3e^(-4t)`

Explanation:

`y''+5y'+4y=0`

Applying the Laplace transform:

`\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0`

With the formulas:

`\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)`

`\mathcal{L}[y']=s\mathcal{L}[y]-y(0)`

`\mathcal{L}[x]=L`

`s^2L-3s+5sL-3+4L=0`

Solving for
`L`

`L(s^2+5s+4)=3s+3`

`L=(3s+3)/(s^2+5s+4)`

`L=(3(s+1))/((s+1)(s+4))`

`L=\frac3{s+4}`

Apply the inverse Laplace transform with this formula:

`\mathcal{L}^(-1)[\frac1{s-a}]=e^(at)`

`y=3\mathcal{L}^(-1)[\frac1{s+4}]=3e^(-4t)`