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

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

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

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

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + \frac{x}{2} + \frac{1}{4}

Solution steps

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

Solve linear ODE:

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + \frac{x}{2} + \frac{1}{4}

y = e^{\frac{x}{2}} (c_{1} cos (\frac{7 x}{2}) + c_{2} sin (\frac{7 x}{2})) + \frac{x}{2} + \frac{1}{4}

2 y^{^{''}} + 2 y^{^{'}} + 2 y = sin (t) + 1

u^{^{''}} + a u^{^{'}} (t) = b

6 y^{^{''}} + 3 y^{^{'}} - 30 y = 16 cos (x)

y^{^{''}} - 5 y^{^{'}} - 36 y = e^{9 x}

y^{^{''}} + y^{^{'}} + y = sin (x) + x + x^{2} - 1

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