What is the solution for y^'+x^2y=1 ?

The solution for y^'+x^2y=1 is y=(integral of e^{(x^3)/3})/(e^{(x^3)/3)}

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y^'+x^2y=1

y^{^{'}} + x^{2} y = 1

y = \frac{\int e ^{\frac{x ^{3}}{3}} d x}{e ^{\frac{x ^{3}}{3}}}

Solution steps

y^{'} (x) + x^{2} y = 1

Solve linear ODE:

y = \frac{\int e ^{\frac{x ^{3}}{3}} d x}{e ^{\frac{x ^{3}}{3}}}

y = \frac{\int e ^{\frac{x ^{3}}{3}} d x}{e ^{\frac{x ^{3}}{3}}}

\frac{- sin ( x ) 2 x}{3}

\frac{d}{d x} (3 x^{9})

\frac{d}{d x} ((ln (x))^{l n (x)})

\int \frac{1}{12 x ^{5}} d x

x \to - \infty + lim (- 5 x^{3})

What is the solution for y^'+x^2y=1 ?
The solution for y^'+x^2y=1 is y=(integral of e^{(x^3)/3})/(e^{(x^3)/3)}