What is the solution for y^{''}+4y^'=10e^t ?

The solution for y^{''}+4y^'=10e^t is y=c_{1}+c_{2}e^{-4t}+2e^t

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y^{''}+4y^'=10e^t

y^{^{''}} + 4 y^{^{'}} = 10 e^{t}

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

Solution steps

y^{''} (t) + 4 y^{'} (t) = 10 e^{t}

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{e ^{4 x} + e ^{- 4 x}}{x})

2 x, x = 3

f^{^{''}} (x) = f (x)

\frac{d y}{d x} + 2 y = 7, y (0) = 1

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

What is the solution for y^{''}+4y^'=10e^t ?
The solution for y^{''}+4y^'=10e^t is y=c_{1}+c_{2}e^{-4t}+2e^t