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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + 5 y = \frac{1}{x} e^{2 x} sin (x)

y^{^{''}} + 4 y = - 16 e^{- 2 x} + 2 sin (x)

y^{^{''}} + 4 y^{^{'}} = 8 x + 1

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

4 y^{^{''}} - y = x e^{\frac{1}{2} x}

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