MCQ
The solution of differential equation $(x^2 -xy)dy = (xy + y^2)dx$ is
- A$xy = ce^{-y/x}$
- ✓$xy = ce^{-x/y}$
- C$yx^2 = ce^{1/x}$
- DNone of these
$\left(x^2-x y\right) d y=\left(x y+y^2\right) d x$
$\Rightarrow \frac{d y}{d x}=\frac{x y+y^2}{x^2-x y}$
which is a homogeneous differential eqn
Puty $= vx$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
So, eqn $(1)$ becomes
$v + x \frac{ dv }{ dx }=\frac{ v + v ^2}{1- v }$
$\Rightarrow x \frac{ dv }{ dx }=\frac{2 v ^2}{1- v }$
$\Rightarrow \frac{1- v }{2 v ^2} dv =\frac{ dx }{ x }$
Integrating both sides, we get
$\Rightarrow \int \frac{1- v }{ v ^2} dv =2 \int \frac{ dx }{ x }$
$\Rightarrow \frac{-1}{ v }-\log v =2 \log x +\log C$
$\Rightarrow \frac{- x }{ y }-\log y +\log x =2 \log x +\log C$
$\Rightarrow \frac{- x }{ y }=\log xy C$
$\Rightarrow ce ^{-x y}=x y \quad$ where $c=\frac{1}{C}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$g(x)=\left\{\begin{array}{cl}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\1, & x=-1\end{array} \text { and } h(x)=2[x]-f(x),\right.$
where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim _{x \rightarrow 1} g(h(x-1))$ is