JEE Main 202225 Jul, Morning shiftMathematicsMediumDifferential Equations163× since 2002The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0(x−y2)dx+y(5x+y2)dy=0 is :AOption A. (y2+x)4=C∣(y2+2x)3∣\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|(y2+x)4=C(y2+2x)3BOption B. (y2+2x)4=C∣(y2+x)3∣\left(y^{2}+2 x\right)^{4}=C\left|\left(y^{2}+x\right)^{3}\right|(y2+2x)4=C(y2+x)3COption C. ∣(y2+x)3∣=C(2y2+x)4\left|\left(y^{2}+x\right)^{3}\right|=\mathrm{C}\left(2 y^{2}+x\right)^{4}(y2+x)3=C(2y2+x)4DOption D. ∣(y2+2x)3∣=C(2y2+x)4\left|\left(y^{2}+2 x\right)^{3}\right|=C\left(2 y^{2}+x\right)^{4}(y2+2x)3=C(2y2+x)4CheckSkip · show solution
