The solution of the equation dydx=x2y-x is:

dydx=x2y-xPut y=vx⇒v+xdvdx=dydxv+xdvdx=x2v−x=12v−1xdvdx=12v−1−v=1−2v2+v2v−1=−(v−1)(2v+1)2v−1(2v−1)(2v+1)(v−1)=−dxx;13(v−1)+43(2v+1)=−dxx13log⁡(v−1)+43⋅12log⁡(2v+1)=log⁡1x+log⁡clog⁡(v−1)i/3+log⁡(2v+1)13/3=log⁡cx=(v−1)1/3(2v+1)2/3=cxy−xx2y+xx2=c3x3⇒ (x−y)(x+2y)2=c.

The solution of the equation dydx=x2y-x is:

dydx=x2y-xPut y=vx⇒v+xdvdx=dydxv+xdvdx=x2v−x=12v−1xdvdx=12v−1−v=1−2v2+v2v−1=−(v−1)(2v+1)2v−1(2v−1)(2v+1)(v−1)=−dxx;13(v−1)+43(2v+1)=−dxx13log⁡(v−1)+43⋅12log⁡(2v+1)=log⁡1x+log⁡clog⁡(v−1)i/3+log⁡(2v+1)13/3=log⁡cx=(v−1)1/3(2v+1)2/3=cxy−xx2y+xx2=c3x3⇒ (x−y)(x+2y)2=c.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The solution of the equation $\frac{d y}{d x} = \frac{x}{2 y - x}$ is:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(x - y) (x + 2 y)^{2} = c$

$y = \frac{x}{2 y - x} + c$

$y = x + c$

$y = (2 y - x) + c$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{d y}{d x} = \frac{x}{2 y - x}$

Put $y = v x \Rightarrow v + x \frac{d v}{d x} = \frac{d y}{d x}$

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{x}{2 v - x} = \frac{1}{2 v - 1} \\ x \frac{d v}{d x} = \frac{1}{2 v - 1} - v = \frac{1 - 2 v^{2} + v}{2 v - 1} = - \frac{(v - 1) (2 v + 1)}{2 v - 1} \\ \frac{(2 v - 1)}{(2 v + 1) (v - 1)} = \frac{- d x}{x}; \frac{1}{3 (v - 1)} + \frac{4}{3 (2 v + 1)} = \frac{- d x}{x} \\ \frac{1}{3} \log (v - 1) + \frac{4}{3} \cdot \frac{1}{2} \log (2 v + 1) = \log \frac{1}{x} + \log c \end{array}$

$\begin{array}{l} \log (v - 1)^{i / 3} + \log (2 v + 1)^{\frac{1}{3} / 3} = \log \frac{c}{x} \\ = (v - 1)^{1 / 3} (2 v + 1)^{2 / 3} = \frac{c}{x} \\ (\frac{y - x}{x}) {(\frac{2 y + x}{x})}^{2} = \frac{c^{3}}{x^{3}} \\ \Rightarrow (x - y) (x + 2 y)^{2} = c . \end{array}$