Answer (1 of 2) 2xy dx (x^2 1) dy =0 => 2xydx = (x^2–1)dy => 2x/(x^2–1)dx = dy/y after integrating both sides => ln(x^2–1) = ln(y) constant => ln(x(xy^22x^2y^3)dx(x^2yx^3y^2)dy=0 (y2xy^2)dx=(x^2yx)dy y(12xy)dx=x(xy1)dy Substitute t=yx and t'=yxy' 3\int \frac {dx} x=\int \frac {(t1)dt} {t^2} This is an example of an exact DE — that is, it has the form F(x,y)dxG(x,y)dy=0, where \frac{\partial F}{\partial y}=\frac{\partial G}{\partial x} The solution to such a DE is given Solve the differential equation (x^2 y^2)dy/dx = 2xy given that y = 1, x = 1 asked in Differential equations by AmanYadav ( 556k points) differential equations
Solve The Differential Equation X2 1 Dy Dx 2xy 1 X2 1 Studyrankersonline
