What is the solution for y^'+y=2te^{-t}+t^2 ?

The solution for y^'+y=2te^{-t}+t^2 is y=(t^2)/(e^t)+t^2-2(t-1)+(c_{1})/(e^t)

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y^'+y=2te^{-t}+t^2

y^{^{'}} + y = 2 t e^{- t} + t^{2}

y = \frac{t ^{2}}{e ^{t}} + t^{2} - 2 (t - 1) + \frac{c _{1}}{e ^{t}}

Solution steps

y^{'} (t) + y = 2 t e^{- t} + t^{2}

Solve linear ODE:

y = \frac{t ^{2}}{e ^{t}} + t^{2} - 2 (t - 1) + \frac{c _{1}}{e ^{t}}

y = \frac{t ^{2}}{e ^{t}} + t^{2} - 2 (t - 1) + \frac{c _{1}}{e ^{t}}

\frac{d}{d x} ((2 x^{4} - 3 x)^{3})

\int \frac{3}{8} d x

\int \frac{2 x}{( x ^{2} - 1 )} d x

f (x) = \frac{x ^{2}}{4 + x}

y^{^{'}} + 7 y = 9

What is the solution for y^'+y=2te^{-t}+t^2 ?
The solution for y^'+y=2te^{-t}+t^2 is y=(t^2)/(e^t)+t^2-2(t-1)+(c_{1})/(e^t)