Question

The differential equation $\varphi \left(x\right)dy=y\{{\varphi }^{{}^{\prime }}\left(x\right)-y\}dx$ is changed in the form df(x, y)=0. Then $f(x,y)$ is:

• A. $\frac{1}{2}\varphi \left(x\right)+y$
• B. $\frac{1}{y}\varphi \left(x\right)-x$
• C. $\frac{1}{y}\varphi \left(x\right)+x$
• D. $\frac{\varphi \left(x\right)}{y}$

Answer 1

Answer: (B)

$\frac{1}{y}\varphi \left(x\right)-x$

$\varphi \left(x\right)dy=y\{{\varphi }^{{}^{\prime }}\left(x\right)-y\}dx$
$\Rightarrow \varphi \left(x\right)dy-y{\varphi }^{\prime }\left(x\right)=-y^{2}dx$
$\Rightarrow \frac{y{\varphi }^{\prime }\left(x\right)-\varphi \left(x\right)dy}{y^2}=dx$
$\Rightarrow d\left(\frac{\varphi \left(x\right)}{y}\right)=dx$ $\Rightarrow d(\frac{\varphi \left(x\right)}{y}-x)=0$
$\therefore f(x,y)=\frac{1}{y}\varphi \left(x\right)-x$