Question:\ y' - y = 2t * e^(2t), y(0)= 1
Answer:
\frac{dy}{dx} -y = 2t*e^{2t})
transform this to the general form:
\frac{dy}{dx} + p(t)y = A(t)
P(t) = 1
Let \mu = e^{\int P(t)*dt}\ = e^{\int-1*dt} = e^{-t}
multiply both sides by \mu(t)
e^{-t} * (\frac{dy}{dx} -y) = e^{-t} * (2t*e^{2t} )
e^{-t}y' - e^{-t}y = e^{-t}(2t*e^{2t})
(ye^{-t})' = 2t*e'
integrate both sides
\int (y*e^{-t})' = \int (2t*e^{t})'*dt
ye^{-t} = 2\int (t*e^{t})'*dt
let u = t and dv= e^{t}*dt
\int u*dv = u*v - \int v*du
=\int t*e^{t} - \int e^{t}*dt
= t*e^{t}-e^{t}+c
=(t-1)e^{t} +c
ye^{-t} = 2[(t-1)e^{t}+c]
multiply both sides by e^{t}
y=2e^{2t}(t-1) +c*e^{t}
from the question we know y(0)=1
y(0) = 2e^{2(0)}(0-1)+c*e^0
1 = 2(-1)+c
C=3
Substitute c=3 in y=2e^{2t}(t-1)+c*e^{t}
y=2e^{2t}(t-1)+3e^t
So, the final solution is y(t) = 2e^{2t}(t-1)+3e^{t}.
