Solve 1st Order Linear ODE Initial Value Problem

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SUMMARY

The discussion focuses on solving the first-order linear ordinary differential equation (ODE) given by y' - y = 2te^(2t) with the initial condition y(0) = 1. The solution process involves using an integrating factor, μ(t) = e^(-t), to transform the equation into a solvable form. After applying integration techniques, the final solution is derived as y(t) = 2te^(2t) - 2e^(2t) + 2e^(t). The solution is confirmed by substituting the initial condition into the derived equation.

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Punchlinegirl
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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?
 
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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}
 

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