Solve the initial value problem y 3y 2y expt y00y01 Solut

 Solve the initial value problem:  y\'\' - 3y\' + 2y = exp(-t),   y(0)=0,y\'(0)=1. 

Solution

Let x = t - 2

t = x + 2.

So, dy/dx = dy/dt dt/dx = dy/dt * 1 = dy/dt,

and similarly d^2y/dx^2 = d^2y/dt^2.

Hence, the initial value problem transforms to
y\'\' - 3y\' + 2y = e^(-(x+2)) = e^(-2) e^(-x), with y(0) = 0 and y\'(0) = 1.

Now, apply L to both sides:
[s^2 Y(s) - 1s - 0] - 3[s Y(s) - 1] + 2 Y(s) = e^(-2) * 1/(s+1).

Solve for Y(s):
(s^2 - 3s + 2) Y(s) = (s + 3) + e^(-2)/(s+1)
Y(s) = (s+3)/[(s-1)(s-2)] + e^(-2)/[(s+1)(s-1)(s-2)].

Using the partial fractions, we can rewrite this as
Y(s) = [5/(s-2) - 4/(s-1)] + (1/6)e^(-2) [1/(s+1) - 3/(s-1) + 2/(s-2)].

y(x) = [5e^(2x) - 4e^x] + (1/6)e^(-2) [e^(-x) - 3e^x + 2e^(2x)].

Since x = t-2, we have
y(x) = [5e^(2(t-2)) - 4e^(t-2)] + (1/6)e^(-2) [e^(-(t-2)) - 3e^(t-2) + 2e^(2(t-2))].