Question #138837

dy/dx=2x+2xy²/y+2x²y


1
Expert's answer
2020-10-18T17:48:02-0400

dydx=2x+2xy2y+2x2y\frac{dy}{dx}=\frac{2x+2xy^2}{y+2x^2y}


Let us solve this differential equation:


dydx=2x(1+y2)y(1+2x2)\frac{dy}{dx}=\frac{2x(1+y^2)}{y(1+2x^2)}


ydy1+y2=2xdx1+2x2\frac{ydy}{1+y^2}=\frac{2xdx}{1+2x^2} (1+y2>01+y^2>0)


ydy1+y2=2xdx1+2x2\int\frac{ydy}{1+y^2}=\int\frac{2xdx}{1+2x^2}


12d(1+y2)1+y2=12d(1+2x2)1+2x2\frac{1}{2}\int\frac{d(1+y^2)}{1+y^2}=\frac{1}{2}\int\frac{d(1+2x^2)}{1+2x^2}


d(1+y2)1+y2=d(1+2x2)1+2x2\int\frac{d(1+y^2)}{1+y^2}=\int\frac{d(1+2x^2)}{1+2x^2}


ln(1+y2)=ln(1+2x2)+ln(C)\ln(1+y^2)=\ln(1+2x^2)+ln(C)


ln(1+y2)=ln((1+2x2)C)\ln(1+y^2)=\ln((1+2x^2)C)


Therefore, the general solution is the following:


1+y2=C(1+2x2)1+y^2=C(1+2x^2)






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