What is the solution for y^{''}+4y^'=2e^{2x} ?

The solution for y^{''}+4y^'=2e^{2x} is y=c_{1}+c_{2}e^{-4x}+1/6 e^{2x}

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y^{''}+4y^'=2e^{2x}

y^{^{''}} + 4 y^{^{'}} = 2 e^{2 x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{1}{6} e^{2 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) = 2 e^{2 x}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} + \frac{1}{6} e^{2 x}

y = c_{1} + c_{2} e^{- 4 x} + \frac{1}{6} e^{2 x}

\int - k

4 x + \frac{5}{x}

\frac{\partial}{\partial x} (6 - x^{2} + x y - 4 y^{2})

x^{2} + 1, x + 1

\int \frac{1}{( x ^{2} + 49 ) ^{\frac{3}{2}}} d x

What is the solution for y^{''}+4y^'=2e^{2x} ?
The solution for y^{''}+4y^'=2e^{2x} is y=c_{1}+c_{2}e^{-4x}+1/6 e^{2x}