Question #288869

Solve:


x²y²(2ydx + xdy) - (5ydx + 7xdy) =0

1
Expert's answer
2022-01-25T17:02:42-0500

We write the equation as



2x2y3dx+x3y2dy5ydx7xdy=02x^2y^3dx+x^3y^2dy-5ydx-7xdy=0

Let's select the total differential x^3y^3



13d(x3y3)+x2y3dx5ydx7xdy=0.\frac{1}{3} d(x^3y^3)+x^2y^3dx-5ydx-7xdy=0.

Let's select the total differential -5xy



13d(x3y3)5d(xy)+x2y3dx2xdy=0.\frac{1}{3} d(x^3y^3)-5d(xy)+x^2y^3dx-2xdy=0.

Consider the equation



x2y3dx2xdy=0x^2y^3dx-2xdy=0

get



xdx=2dyy3.xdx=\frac{2dy}{y^3}.

As result



x22+1y2=C\frac{x^2}{2}+\frac{1}{y^2}=C

where C is constant.

Hence. the general solution of the equation is



x3y335xy+x22+1y2=c\frac{x^3y^3}{3}-5xy+\frac{x^2}{2}+\frac{1}{y^2}=c






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