What is the solution for y^'=2y+x^2+7 ?

The solution for y^'=2y+x^2+7 is y=-(x^2)/2-x/2-15/4+c_{1}e^{2x}

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y^'=2y+x^2+7

y^{^{'}} = 2 y + x^{2} + 7

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{15}{4} + c_{1} e^{2 x}

Solution steps

y^{'} (x) = 2 y + x^{2} + 7

Solve linear ODE:

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{15}{4} + c_{1} e^{2 x}

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{15}{4} + c_{1} e^{2 x}

\frac{d ^{2}}{d x ^{2}} (e^{3 x})

\frac{1}{s ^{2} + 8} (- \frac{40}{33})

\frac{d}{d x} (- 0.025 x^{2} + 9 x)

\frac{7}{2 7 x + 2}, at a = 2

\int x csc^{2} (x) d x

What is the solution for y^'=2y+x^2+7 ?
The solution for y^'=2y+x^2+7 is y=-(x^2)/2-x/2-15/4+c_{1}e^{2x}