Solve 1st Order Linear ODE Initial Value Problem

[Punchlinegirl]

Find the solution of the given initial value problem
y'-y=2te^2t y(0)=1

so p(t)=-1
then $$\mu (t)= e^/int -1 dt= e^-t$$
e^-t y'-e^-t y=2te^2te^-t
d/dt [e^-t y]=2te^t dt
e^-t =$$\int 2te^t dt$$
e^-t y= 2te^t-2e^t
y=-2(t-1)+ce^-t
plugging in the initial conditon gives me
y(t)=-2(t-1)-e^-t

Am I doing this right? If not, can someone help?

 

[courtrigrad]

$$I = e^{-\int t} = e^{-t dt}$$.

So you get $$ye^{-t} = 2\int te^{t} dt = 2te^{t} - 2e^{t} + C$$

$$y = \frac{2te^{t} - 2e^{t} + C}{e^{-t}}$$

$$y = \frac{2te^{t} - 2e^{t} + 2}{e^{-t}}$$

$$y = 2te^{2t} - 2e^{2t} + 2e^{t}$$