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

The solution for y^{''}+4y=5xe^{2x} is y=c_{1}cos(2x)+c_{2}sin(2x)+(5e^{2x}x)/8-5/16 e^{2x}

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y^{''}+4y=5xe^{2x}

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 e ^{2 x} x}{8} - \frac{5}{16} e^{2 x}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 e ^{2 x} x}{8} - \frac{5}{16} e^{2 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{5 e ^{2 x} x}{8} - \frac{5}{16} e^{2 x}

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 4 x^{2} + 8 x + 10

(D^{3} + D - 10) y = 29 e^{4 x}

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

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

What is the solution for y^{''}+4y=5xe^{2x} ?
The solution for y^{''}+4y=5xe^{2x} is y=c_{1}cos(2x)+c_{2}sin(2x)+(5e^{2x}x)/8-5/16 e^{2x}