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

The solution for y^{''}-4y^'-4y=e^{2x} is y=c_{1}e^{2(1+sqrt(2))x}+c_{2}e^{2(1-sqrt(2))x}-1/8 e^{2x}

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

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

y = c_{1} e^{2 (1 + 2) x} + c_{2} e^{2 (1 - 2) x} - \frac{1}{8} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{2 (1 + 2) x} + c_{2} e^{2 (1 - 2) x} - \frac{1}{8} e^{2 x}

y = c_{1} e^{2 (1 + 2) x} + c_{2} e^{2 (1 - 2) x} - \frac{1}{8} e^{2 x}

y = \frac{x ^{2}}{2} - 4 x + 6, y = \frac{x}{2} - 1, x = 0, x = 5

\int \frac{441}{x ^{3} - 21 x ^{2}} d x

\int \frac{x ^{2}}{( 25 + x ^{2} ) ^{2}} d x

5 x y \frac{d y}{d x} = 1

y^{^{''}} - 3 y^{^{'}} - 12 y = 0

What is the solution for y^{''}-4y^'-4y=e^{2x} ?
The solution for y^{''}-4y^'-4y=e^{2x} is y=c_{1}e^{2(1+sqrt(2))x}+c_{2}e^{2(1-sqrt(2))x}-1/8 e^{2x}