What is the solution for y^'+3y=x^2 ?

The solution for y^'+3y=x^2 is y=(x^2)/3-(2(3x-1))/(27)+(c_{1})/(e^{3x)}

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y^'+3y=x^2

y^{^{'}} + 3 y = x^{2}

y = \frac{x ^{2}}{3} - \frac{2 ( 3 x - 1 )}{27} + \frac{c _{1}}{e ^{3 x}}

Solution steps

y^{'} (x) + 3 y = x^{2}

Solve linear ODE:

y = \frac{x ^{2}}{3} - \frac{2 ( 3 x - 1 )}{27} + \frac{c _{1}}{e ^{3 x}}

y = \frac{x ^{2}}{3} - \frac{2 ( 3 x - 1 )}{27} + \frac{c _{1}}{e ^{3 x}}

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

\frac{d y}{d x} + 4 e^{x + y} = 0

y = x^{2} - 9, y = (2 x - 1) (x + 3)

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

\frac{d}{d x} (9 (3 x + 5)^{2})

What is the solution for y^'+3y=x^2 ?
The solution for y^'+3y=x^2 is y=(x^2)/3-(2(3x-1))/(27)+(c_{1})/(e^{3x)}