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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = sec (t)

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 1, y^{^{'}} (0) = 0, y (0) = 0

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

y^{^{''''}} - y = e^{2 t}

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