JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q11

MathematicsSingle correct+4 / -1Question 11 of 90

The general solution of the differential equation $\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$ is :

\left(y^2+x\right)^4=C\left|\left(y^2+2x\right)^3\right|

\left(y^2+2x\right)^4=C\left|\left(y^2+x\right)^3\right|

\left|\left(y^2+x\right)^3\right|=C\left(2y^2+x\right)^4

\left|\left(y^2+2x\right)^3\right|=C\left(2y^2+x\right)^4

Answer and worked solution

Official answer: A

Substitute $u=y^2$ , then $u=vx$ : $\dfrac{dx}x=-\left(\dfrac4{v+1}-\dfrac3{v+2}\right)dv$ , so $x(v+1)^4=C(v+2)^3$ .

Back. $(y^2+x)^4=C\left|(y^2+2x)^3\right|$ .

(A)

Topics:Homogeneous equations

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 11