What is the solution for y^{''}+4y=10e^{5x} ?

The solution for y^{''}+4y=10e^{5x} is y=c_{1}cos(2x)+c_{2}sin(2x)+10/29 e^{5x}

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y^{''}+4y=10e^{5x}

y^{^{''}} + 4 y = 10 e^{5 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{10}{29} e^{5 x}

Solution steps

y^{''} (x) + 4 y = 10 e^{5 x}

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{10}{29} e^{5 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{10}{29} e^{5 x}

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = sec (t)

y^{^{''}} + y = cos (x) - sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 1, y^{^{'}} (0) = 0, y (0) = 0

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

What is the solution for y^{''}+4y=10e^{5x} ?
The solution for y^{''}+4y=10e^{5x} is y=c_{1}cos(2x)+c_{2}sin(2x)+10/29 e^{5x}