What is the solution for y^{''}-3y^'-4y=3e^{2x} ?

The solution for y^{''}-3y^'-4y=3e^{2x} is y=c_{1}e^{4x}+c_{2}e^{-x}-1/2 e^{2x}

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y^{''}-3y^'-4y=3e^{2x}

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2 x}

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

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = 3 e^{2 x}

Solve linear ODE:

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

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

\int_{2}^{3} 3 x d x

\int_{0}^{\frac{π}{4}} (tan (x))^{2} d x

\frac{d}{d x} (- 6 x^{3})

x \to - 1 lim (3 f (x) - 2 h (x))

ln (\frac{x + 7}{x - 7})

What is the solution for y^{''}-3y^'-4y=3e^{2x} ?
The solution for y^{''}-3y^'-4y=3e^{2x} is y=c_{1}e^{4x}+c_{2}e^{-x}-1/2 e^{2x}