What is the solution for y^{''}-4y^'+3y=xe^x ?

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

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

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

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

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = x e^{x}

Solve linear ODE:

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3} t ln (t)

y^{^{''}} + y = cos (2 t) cos (t)

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{t} + t, y^{^{'}} (0) = 0, y (0) = 1

y^{^{''}} - y = 2 e^{x} - 1 - 6 \cdot x \cdot e^{x}

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