What is the solution for y^'=2y+x^2+9 ?

The solution for y^'=2y+x^2+9 is y=-(x^2)/2-x/2-19/4+c_{1}e^{2x}

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y^'=2y+x^2+9

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

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{19}{4} + c_{1} e^{2 x}

Solution steps

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

Solve linear ODE:

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{19}{4} + c_{1} e^{2 x}

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{19}{4} + c_{1} e^{2 x}

\frac{d}{d x} ((x + 1)^{x^{2}})

f (x) = x^{2} + a x + 2, at x = 1

\int - 6 ln (x^{2} - 1) d x

\int (x cos (4 x)) d x

\frac{d}{d x} (\frac{2 x + 2 x ^{2}}{( 1 + 2 x ) ^{2}})

What is the solution for y^'=2y+x^2+9 ?
The solution for y^'=2y+x^2+9 is y=-(x^2)/2-x/2-19/4+c_{1}e^{2x}