Solution of the differential equation dydx+(xy)(x2+y2−12(x2+y2)+1)=0 is[c is arbitrary constant]

ydyxdx+(x2+y2−12(x2+y2)+1)=0Let u=x2, v=y2 dvdu+(u+v−12(u+v)+1)=0, let t=u+v⇒dtdu−1+t−12t+1=0⇒dtdu=t+22t+1 dtdu=t+22t+1⇒∫(2−3t+2)dt⇒2t−3ln(t+2)=u+c

Solution of the differential equation dydx+(xy)(x2+y2−12(x2+y2)+1)=0 is[c is arbitrary constant]

ydyxdx+(x2+y2−12(x2+y2)+1)=0Let u=x2, v=y2 dvdu+(u+v−12(u+v)+1)=0, let t=u+v⇒dtdu−1+t−12t+1=0⇒dtdu=t+22t+1 dtdu=t+22t+1⇒∫(2−3t+2)dt⇒2t−3ln(t+2)=u+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

Solution of the differential equation $\frac{d y}{d x} + (\frac{x}{y}) (\frac{x^{2} + y^{2} - 1}{2 (x^{2} + y^{2}) + 1}) = 0$ is
[c is arbitrary constant]

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

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

$x^{2} + 2 x y - 3 l n (x^{2} + y^{2} + 2) = c$

$- x^{2} - 2 y^{2} - 3 l n (x^{2} + y^{2}) = c$

$x^{2} + 2 y^{2} - 3 l n (x^{2} + y^{2} + 2) = c$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \frac{y d y}{x d x} + (\frac{x^{2} + y^{2} - 1}{2 (x^{2} + y^{2}) + 1}) = 0 \\ L e t u = x^{2}, v = y^{2} \end{array} \begin{array}{l} \frac{d v}{d u} + (\frac{u + v - 1}{2 (u + v) + 1}) = 0, l e t t = u + v \\ \Rightarrow \frac{d t}{d u} - 1 + \frac{t - 1}{2 t + 1} = 0 \Rightarrow \frac{d t}{d u} = \frac{t + 2}{2 t + 1} \end{array} \begin{array}{l} \frac{d t}{d u} = \frac{t + 2}{2 t + 1} \Rightarrow \int (2 - \frac{3}{t + 2}) d t \Rightarrow 2 t - 3 \ln (t + 2) = u + c \end{array}$