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

The solution for y^{''}-y^'+2y=4e^x is y=e^{x/2}(c_{1}cos((sqrt(7)x)/2)+c_{2}sin((sqrt(7)x)/2))+2e^x

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

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

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + 2 e^{x}

Solution steps

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

Solve linear ODE:

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + 2 e^{x}

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + 2 e^{x}

y^{^{''}} - 3 y^{^{'}} - 6 y = 2 e^{2 x} + 4

(4 d^{4} + 20 d^{3} + 35 d^{2} + 25 d + 6) y = o

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 10 cos (2 t)

y^{^{''}} - 7 y^{^{'}} + 12 y = e^{x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 1 + 3 x + x^{2}

What is the solution for y^{''}-y^'+2y=4e^x ?
The solution for y^{''}-y^'+2y=4e^x is y=e^{x/2}(c_{1}cos((sqrt(7)x)/2)+c_{2}sin((sqrt(7)x)/2))+2e^x