0 votes 0 votes The general solution of the differential equation $2y{y}'-x=0$ is (assuming $C$ as an arbitrary constant of integration) $x^{2}-y^{2}=C$ $2x^{2}-y^{2}=C$ $2y^{2}-x^{2}=C$ $x^{2}+y^{2}=C$ Calculus isi2016-dcg calculus differential-equation non-gate + – gatecse asked Sep 18, 2019 recategorized Nov 19, 2019 by Lakshman Bhaiya gatecse 260 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Rearranging the terms of the differential equation, $2y\frac{dy}{dx}=x$. Now integrating on both sides, $\Rightarrow$ $\int 2ydy=\int xdx$ $\Rightarrow$ $2\times (\frac{y^{2}}{2})= \frac{x^{2}}{2} +k$ i.e. $2y^{2}-x^{2}=C$ Option C is the answer. haralk10 answered Mar 14, 2020 edited Apr 12, 2020 by haralk10 haralk10 comment Share Follow See all 2 Comments See all 2 2 Comments reply thebiggercypher commented Apr 12, 2020 reply Follow Share Your answer is wrong . Option C is correct . Integration of x will be $x^2/2$ Correct answer is $2y^2-x^2=C$ 1 votes 1 votes haralk10 commented Apr 12, 2020 reply Follow Share Corrected my mistake. Thank you for pointing out 0 votes 0 votes Please log in or register to add a comment.