What is the solution for y^{''}-4y=e^{2t} ?

The solution for y^{''}-4y=e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{2t}t)/4

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y^{''}-4y=e^{2t}

y^{^{''}} - 4 y = e^{2 t}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

Solution steps

y^{''} (t) - 4 y = e^{2 t}

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{e ^{2 t} t}{4}

\int_{\frac{1}{2}}^{3} 8 x ln (2 x) d x

t \to 0 lim (\frac{( tan ( t ) sec ( 2 t ) )}{3 t})

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

ln (x)

\frac{d}{d x} ((x^{2} + y^{2} + z^{2})^{\frac{1}{2}})

What is the solution for y^{''}-4y=e^{2t} ?
The solution for y^{''}-4y=e^{2t} is y=c_{1}e^{2t}+c_{2}e^{-2t}+(e^{2t}t)/4