Question

(d^2 y)/(dt^2) - 2 dy/dt + 1=0 with y(t=0)=5 and y ̇(t=0)=-9

use laplace transform

Answer 1

Let's take the transform of every term in the differential equation

$L[y'']-L[2y']+L[1]=0$

It is known, that

$L[y'']=s^2L[y]-sy(0)-y'(0)=s^2Y(s)-5s+9$

$L[y']=sY(s)-y(0)=sY(s)-5$ .

Here $Y(s)=L[y]$

The Laplace-transformed differential equation is

$s^2Y(s)-5s+9-2(sY(s)-5)+\frac{1}{s}=0$

$Y(s^2-2s)=5s-\frac{1}{s}-19$

$Y=\frac{5s^2-19s-1}{s(s^2-2s)}=\frac{1}{2s^2}+\frac{39}{4s}-\frac{19}{4(s-2)}$

Recalling the inverse transform will lead us to

$\frac{1}{s^2}\gets x;\quad \frac{1}{s}\gets 1;\quad \frac{1}{s-2}\gets e^{2t}$

Therefore

$Y\gets y=\frac{t}{2}+\frac{39}{4}-\frac{19}{4}e^{2t}$