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

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

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y^{''}-3y^'+2y=e^{3x}(1+x)

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (1 + x)

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

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = e^{3 x} (1 + x)

Solve linear ODE:

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

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

y = 4^{x} (2 x^{4})

3 (e^{x} + e^{x} x)

\frac{d}{d θ} (6 cos (θ) + 8 sin (θ))

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

\frac{d y}{d x} + \frac{8 y}{x + 2} = (x + 2)^{6}

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