What is the solution for y^'=y+3te^t,y(0)=4 ?

The solution for y^'=y+3te^t,y(0)=4 is y=(3t^2)/(2e^{-t)}+4e^t

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y^'=y+3te^t,y(0)=4

y^{^{'}} = y + 3 t e^{t}, y (0) = 4

y = \frac{3 t ^{2}}{2 e ^{- t}} + 4 e^{t}

Solution steps

y^{'} (t) = y + 3 t e^{t}

Solve linear ODE:

y = \frac{3 t ^{2}}{2 e ^{- t}} + 4 e^{t}

y = \frac{3 t ^{2}}{2 e ^{- t}} + 4 e^{t}

x \to 3 lim (x - 8)

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

y^{^{''}} - 16 y = 2 e^{4 x}

\int_{1}^{2} t ln (t) d t

\frac{d}{d x} ((5 x - 4) (2 x^{3} - x^{2} + 1))

What is the solution for y^'=y+3te^t,y(0)=4 ?
The solution for y^'=y+3te^t,y(0)=4 is y=(3t^2)/(2e^{-t)}+4e^t