What is the solution for 2x^2y+x^3y^'=1 ?

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

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2x^2y+x^3y^'=1

2 x^{2} y + x^{3} y^{^{'}} = 1

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

Solution steps

2 x^{2} y + x^{3} y^{'} (x) = 1

Solve linear ODE:

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

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

\int \frac{16}{64 + ( 2 π x ) ^{2}} d x

y^{^{'}} (x^{2} + 1) + 3 x y = 6 x

3 cos^{2} (x)

\frac{d}{d x} (2 sin (x) + cos (x))

(x - 2)^{^{'}}

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