What is the solution for y^{''}+y=e^x(6-x) ?

The solution for y^{''}+y=e^x(6-x) is y=c_{1}cos(x)+c_{2}sin(x)-(e^xx)/2+7/2 e^x

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y^{''}+y=e^x(6-x)

y^{^{''}} + y = e^{x} (6 - x)

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

Solution steps

y^{''} (x) + y = e^{x} (6 - x)

Solve linear ODE:

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

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

x^{^{''}} + 6 x^{^{'}} + 9 x = 10 cos (2 t)

y^{^{'''}} + 2 y^{^{'}} = 3 (y^{^{''}} - x^{2} + 2)

y^{^{''}} - 4 y^{^{'}} + 3 y = 3 + 3 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{'''}} - 3 y^{^{'}} - 2 y = e^{x} (x e^{x} + 1)

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

What is the solution for y^{''}+y=e^x(6-x) ?
The solution for y^{''}+y=e^x(6-x) is y=c_{1}cos(x)+c_{2}sin(x)-(e^xx)/2+7/2 e^x