What is the solution for y^{''}+4y=5(e^x+e^{3x}) ?

The solution for y^{''}+4y=5(e^x+e^{3x}) is y=c_{1}cos(2x)+c_{2}sin(2x)+e^x+5/13 e^{3x}

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y^{''}+4y=5(e^x+e^{3x})

y^{^{''}} + 4 y = 5 (e^{x} + e^{3 x})

y = c_{1} cos (2 x) + c_{2} sin (2 x) + e^{x} + \frac{5}{13} e^{3 x}

Solution steps

y^{''} (x) + 4 y = 5 (e^{x} + e^{3 x})

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + e^{x} + \frac{5}{13} e^{3 x}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + e^{x} + \frac{5}{13} e^{3 x}

\frac{d ^{2}}{d x ^{2}} (θ (x)) - (a + b) θ (x) = (a \cdot c + b)

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

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

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

d, d (q) = \frac{300}{0.4 q + 8}

What is the solution for y^{''}+4y=5(e^x+e^{3x}) ?
The solution for y^{''}+4y=5(e^x+e^{3x}) is y=c_{1}cos(2x)+c_{2}sin(2x)+e^x+5/13 e^{3x}