What is the solution for y^'=2y+x^2+5 ?

The solution for y^'=2y+x^2+5 is y=-(x^2)/2-x/2-11/4+c_{1}e^{2x}

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y^'=2y+x^2+5

y^{^{'}} = 2 y + x^{2} + 5

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{11}{4} + c_{1} e^{2 x}

Solution steps

y^{'} (x) = 2 y + x^{2} + 5

Solve linear ODE:

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{11}{4} + c_{1} e^{2 x}

y = - \frac{x ^{2}}{2} - \frac{x}{2} - \frac{11}{4} + c_{1} e^{2 x}

\int_{0}^{π} 3 x sin (x) d x

\frac{d}{d x} (3 sec (6 x))

\frac{10}{s ( s + 2 ) ^{2}}

\int_{0}^{\frac{π}{2}} 3 sin^{2} (2 x) d x

f (x) = 3 x^{2} - x

What is the solution for y^'=2y+x^2+5 ?
The solution for y^'=2y+x^2+5 is y=-(x^2)/2-x/2-11/4+c_{1}e^{2x}