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

The solution for y^{''}-2y^'+y=xe^x+4 is y=c_{1}e^x+c_{2}xe^x+(e^xx^3)/6+4

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{3}}{6} + 4

\int (x^{2} (x^{3} - 1)^{10}) d x

\frac{d}{d x} (\frac{e ^{2}}{x ( 3 x - 5 )})

\int \frac{3 x ^{2} + 5 x + 3}{( x ^{2} + 1 ) ^{2}} d x

f (x) = 5 x^{2} tan (1 - x)

x (x^{2} + 7)^{4}

What is the solution for y^{''}-2y^'+y=xe^x+4 ?
The solution for y^{''}-2y^'+y=xe^x+4 is y=c_{1}e^x+c_{2}xe^x+(e^xx^3)/6+4