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}

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