What is the solution for 2y^'+y=3t^2 ?

The solution for 2y^'+y=3t^2 is y=3(t^2-8(1/2 t-1))+(c_{1})/(e^{t/2)}

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2y^'+y=3t^2

2 y^{^{'}} + y = 3 t^{2}

y = 3 (t^{2} - 8 (\frac{1}{2} t - 1)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

Solution steps

2 y^{'} (t) + y = 3 t^{2}

Solve linear ODE:

y = 3 (t^{2} - 8 (\frac{1}{2} t - 1)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

y = 3 (t^{2} - 8 (\frac{1}{2} t - 1)) + \frac{c _{1}}{e ^{\frac{t}{2}}}

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

\frac{\partial}{\partial v} (\frac{u}{v})

\frac{d}{d x} (5 + x)

\frac{1}{2} (e^{x} + e^{- x})

\int sin (x + y) d x

What is the solution for 2y^'+y=3t^2 ?
The solution for 2y^'+y=3t^2 is y=3(t^2-8(1/2 t-1))+(c_{1})/(e^{t/2)}