What is the solution for xy^'+y=x^2ln(x) ?

The solution for xy^'+y=x^2ln(x) is y=(x^2ln(x))/3-(x^2)/9+(c_{1})/x

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xy^'+y=x^2ln(x)

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

y = \frac{x ^{2} ln ( x )}{3} - \frac{x ^{2}}{9} + \frac{c _{1}}{x}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2} ln ( x )}{3} - \frac{x ^{2}}{9} + \frac{c _{1}}{x}

y = \frac{x ^{2} ln ( x )}{3} - \frac{x ^{2}}{9} + \frac{c _{1}}{x}

n = 1 \sum \infty (0.5)^{- n}

\int 4 x cos (3 x) d x

\frac{\partial}{\partial u} (u^{2} + v^{2})

\int \frac{6 ln ( 4 x )}{5} d x

- 2 sin (x)

What is the solution for xy^'+y=x^2ln(x) ?
The solution for xy^'+y=x^2ln(x) is y=(x^2ln(x))/3-(x^2)/9+(c_{1})/x