What is the solution for 4y^{''}+25y=41e^{2x} ?

The solution for 4y^{''}+25y=41e^{2x} is y=c_{1}cos((5x)/2)+c_{2}sin((5x)/2)+e^{2x}

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4y^{''}+25y=41e^{2x}

4 y^{^{''}} + 25 y = 41 e^{2 x}

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

Solution steps

4 y^{''} (x) + 25 y = 41 e^{2 x}

Solve linear ODE:

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

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

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

y^{^{''}} - 8 y^{^{'}} + 2 y = t e^{t}

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

y^{^{''}} - 5 y^{^{'}} + 6 y = 9 x + 2 e^{x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 26 x + 47

What is the solution for 4y^{''}+25y=41e^{2x} ?
The solution for 4y^{''}+25y=41e^{2x} is y=c_{1}cos((5x)/2)+c_{2}sin((5x)/2)+e^{2x}