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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(d^{2} + 5 \cdot d + 6) y = (d^{2} + 7 \cdot d + 11) \cdot (t)

(d^{2} - 8 d + 9) y = 8 sin^{5} (x)

(d^{2} + 5 d + 6) \cdot y = x^{2}

(d^{3} + 8) y = x^{4} + 2 x + 1

d \cdot e (d^{2} - 4) y = (e^{2})^{x}

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