What is the solution for y^{''}-3y^'-4y=3x^2+2 ?

The solution for y^{''}-3y^'-4y=3x^2+2 is y=c_{1}e^{4x}+c_{2}e^{-x}-(3x^2)/4+(9x)/8-55/32

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y^{''}-3y^'-4y=3x^2+2

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 x^{2} + 2

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 x ^{2}}{4} + \frac{9 x}{8} - \frac{55}{32}

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = 3 x^{2} + 2

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 x ^{2}}{4} + \frac{9 x}{8} - \frac{55}{32}

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{3 x ^{2}}{4} + \frac{9 x}{8} - \frac{55}{32}

\int \frac{a + b x ^{6}}{7 a x + b x ^{7}} d x

4 e^{- 5 t} - 3 e^{- t}

\frac{d y}{d t} + y = sin (t)

t sin (3 t) - 2 t cos (t)

2 sin (x^{2})

What is the solution for y^{''}-3y^'-4y=3x^2+2 ?
The solution for y^{''}-3y^'-4y=3x^2+2 is y=c_{1}e^{4x}+c_{2}e^{-x}-(3x^2)/4+(9x)/8-55/32