(1 + 𝑥^2) (xdy+ydx) =-2yx^2
The given DE is not solvable, we put dx on right side.
(1+𝑥2)(xdy+ydx)=−2yx2dx⇒(1+x2)(d(xy))=−2yx2dx(1 + 𝑥^2) (xdy+ydx) =-2yx^2 dx \\ \Rightarrow (1+x^2)(d(xy))=-2yx^2dx(1+x2)(xdy+ydx)=−2yx2dx⇒(1+x2)(d(xy))=−2yx2dx
⇒d(xy)=−2yx21+x2dx⇒d(xy)xy=−2x1+x2dx\Rightarrow d(xy)=-\dfrac{2yx^2}{1+x^2} dx \\ \Rightarrow \dfrac{d(xy)}{xy}=-\dfrac{2x}{1+x^2}dx⇒d(xy)=−1+x22yx2dx⇒xyd(xy)=−1+x22xdx
On integrating both sides,
ln(xy)=−ln(1+x2)+lnC⇒ln(xy(1+x2))=lnC⇒xy(1+x2)=C\ln (xy)=-\ln(1+x^2)+\ln C \\ \Rightarrow \ln (xy(1+x^2))=\ln C \\ \Rightarrow xy(1+x^2)= Cln(xy)=−ln(1+x2)+lnC⇒ln(xy(1+x2))=lnC⇒xy(1+x2)=C
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