The D.E whose solution is xy=ax2+bx is

Given xy=ax2+bxdiff w.r. to 'x'⇒ xy1+y=2ax−bx2diff w.r. to 'x'⇒xy2+y1+y1=2a+2bx3⇒ x3y2+2y1x2=2(ax2+bx)⇒ x3y2+2y1x2=2xy⇒ x2y2+2xy1=2y

The D.E whose solution is xy=ax2+bx is

Given xy=ax2+bxdiff w.r. to 'x'⇒ xy1+y=2ax−bx2diff w.r. to 'x'⇒xy2+y1+y1=2a+2bx3⇒ x3y2+2y1x2=2(ax2+bx)⇒ x3y2+2y1x2=2xy⇒ x2y2+2xy1=2y

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 D.E whose solution is $xy = {ax}^{2} + \frac{b}{x}$ 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

$x^{2} y_{2} - {xy}_{1} + 2 y = 0$

$x^{2} y_{2} + 2 {xy}_{1} = 2 y$

$x^{2} y_{2} + {xy}_{1} + 2 y = 0$

$x^{2} y_{2} + {xy}_{1} + y = 0$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} G i v e n x y = a x^{2} + \frac{b}{x} \\ d i f f w . r . t o' x' \\ \Rightarrow x y_{1} + y = 2 a x - \frac{b}{x^{2}} \\ d i f f w . r . t o' x' \\ \Rightarrow x y_{2} + y_{1} + y_{1} = 2 a + \frac{2 b}{x^{3}} \\ \Rightarrow x^{3} y_{2} + 2 y_{1} x^{2} = 2 (a x^{2} + \frac{b}{x}) \\ \Rightarrow x^{3} y_{2} + 2 y_{1} x^{2} = 2 x y \\ \Rightarrow x^{2} y_{2} + 2 x y_{1} = 2 y \end{array}$