What is the solution for y^{''}-5y^'+6y=e^{2x}+4xe^x ?

The solution for y^{''}-5y^'+6y=e^{2x}+4xe^x is y=c_{1}e^{3x}+c_{2}e^{2x}-e^{2x}x+2xe^x+3e^x

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y^{''}-5y^'+6y=e^{2x}+4xe^x

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{2 x} + 4 x e^{x}

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = e^{2 x} + 4 x e^{x}

Solve linear ODE:

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

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

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

y^{^{'''}} - 2 y^{^{''}} + 4 y^{^{'}} - 8 y = 8 sin (2 x)

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = x ln (x)

y^{^{''}} + 4 y = 16 t, y (0) = 3

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

What is the solution for y^{''}-5y^'+6y=e^{2x}+4xe^x ?
The solution for y^{''}-5y^'+6y=e^{2x}+4xe^x is y=c_{1}e^{3x}+c_{2}e^{2x}-e^{2x}x+2xe^x+3e^x