What is the solution for y^{''}+4y+4=-5e^{2x} ?

The solution for y^{''}+4y+4=-5e^{2x} is y=c_{1}cos(2x)+c_{2}sin(2x)-5/8 e^{2x}-1

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y^{''}+4y+4=-5e^{2x}

y^{^{''}} + 4 y + 4 = - 5 e^{2 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{5}{8} e^{2 x} - 1

Solution steps

y^{''} (x) + 4 y + 4 = - 5 e^{2 x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{5}{8} e^{2 x} - 1

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{5}{8} e^{2 x} - 1

What is the solution for y^{''}+4y+4=-5e^{2x} ?
The solution for y^{''}+4y+4=-5e^{2x} is y=c_{1}cos(2x)+c_{2}sin(2x)-5/8 e^{2x}-1