What is the solution for y^{''}+25y=82e^{4x} ?

The solution for y^{''}+25y=82e^{4x} is y=c_{1}cos(5x)+c_{2}sin(5x)+2e^{4x}

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y^{''}+25y=82e^{4x}

y^{^{''}} + 25 y = 82 e^{4 x}

y = c_{1} cos (5 x) + c_{2} sin (5 x) + 2 e^{4 x}

Solution steps

y^{''} (x) + 25 y = 82 e^{4 x}

Solve linear ODE:

y = c_{1} cos (5 x) + c_{2} sin (5 x) + 2 e^{4 x}

y = c_{1} cos (5 x) + c_{2} sin (5 x) + 2 e^{4 x}

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

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t - 3)

y^{^{''}} - 6 y^{^{'}} + 9 y = 2, y^{^{'}} (0) = 6

y^{^{''}} - 2 y^{^{'}} - 15 y = - 37 t + 30 t^{2}

y^{^{''}} - y^{^{'}} - 6 y = 4 cos (3 x)

What is the solution for y^{''}+25y=82e^{4x} ?
The solution for y^{''}+25y=82e^{4x} is y=c_{1}cos(5x)+c_{2}sin(5x)+2e^{4x}