What is the solution for y^{''}-2y^'=9x ?

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

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y^{''}-2y^'=9x

y^{^{''}} - 2 y^{^{'}} = 9 x

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

Solution steps

y^{''} (x) - 2 y^{'} (x) = 9 x

Solve linear ODE:

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

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

\int_{0}^{2} (3 x^{2} - 2) d x

d er i v a t i v e (2 + u^{2})^{5} (5 - 9 u^{2})^{8}

t an g e n t f (x) = 5 + 4 x^{2} - 2 x^{3}, (1, 7)

\int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} 4 x ^{3}}{x ^{2}} d x

\frac{d}{d x} (\frac{sin ( x )}{1 + cos ( x )})

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