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}

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}