The solution of the differential equation dydx=−x2+3y23x2+y2,y(1)=0 is

y=vx,dydx=v+xdvdx v+xdvdx=−1+3v23+v2xdvdx=−(v+1)33+v2∴−∫dxx=∫v2+3(v+1)3dv=∫u2−2u+uu3du , where v+1=u−ln⁡x=ln⁡(v+1)+2v+1−2(v+1)2 (∵y=0,x=1⇒v=0⇒c=0) ∴ln⁡(x+y)+2xy(x+y)2=0

The solution of the differential equation dydx=−x2+3y23x2+y2,y(1)=0 is

y=vx,dydx=v+xdvdx v+xdvdx=−1+3v23+v2xdvdx=−(v+1)33+v2∴−∫dxx=∫v2+3(v+1)3dv=∫u2−2u+uu3du , where v+1=u−ln⁡x=ln⁡(v+1)+2v+1−2(v+1)2 (∵y=0,x=1⇒v=0⇒c=0) ∴ln⁡(x+y)+2xy(x+y)2=0

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 differential equation $\frac{dy}{dx} = - (\frac{x^{2} + 3 y^{2}}{3 x^{2} + y^{2}}), y (1) = 0$ 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

$\log_{e} | x + y | + \frac{2 xy}{(x + y)^{2}} = 0$

$\log_{e} | x + y | + \frac{xy}{(x + y)^{2}} = 0$

$\log_{e} | x + y | - \frac{2 xy}{(x + y)^{2}} = 0$

$\log_{e} | x + y | - \frac{xy}{(x + y)^{2}} = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} y = vx, \frac{dy}{dx} = v + x \frac{dv}{dx} \begin{array}{l} v + x \frac{dv}{dx} = - (\frac{1 + 3 v^{2}}{3 + v^{2}}) \\ x \frac{dv}{dx} = - \frac{(v + 1)^{3}}{3 + v^{2}} \end{array} \\ ∴ - \int \frac{dx}{x} = \int \frac{v^{2} + 3}{(v + 1)^{3}} dv = \int \frac{u^{2} - 2 u + u}{u^{3}} du, where v + 1 = u \\ - \ln x = \ln (v + 1) + \frac{2}{v + 1} - \frac{2}{(v + 1)^{2}} (∵ y = 0, x = 1 \Rightarrow v = 0 \Rightarrow c = 0) ∴ \ln (x + y) + \frac{2 xy}{(x + y)^{2}} = 0 \end{array}$