Question

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

Answer 1

Answer:

$y=3e^{-4t}$

Step-by-step 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 $L$

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

$L=\frac{3s+3}{s^2+5s+4}$

$L=\frac{3(s+1)}{(s+1)(s+4)}$

$L=\frac3{s+4}$

Apply the inverse Laplace transform with this formula:

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

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