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Q.
If the solution of the differential equation $\frac{ dy }{ dx }=\frac{ ax +3}{2 y + f }$ represents a circle, then the value of $'a'$ is
Differential Equations
Solution:
We have,
$\frac{ dy }{ dx }=\frac{ ax +3}{2 y + f }$
$ \Rightarrow ( ax +3) dx =(2 y + f ) dy $
$\Rightarrow a \frac{ x ^{2}}{2}+3 x = y ^{2}+ fy + C$ (Integrating)
$\Rightarrow -\frac{ a }{2} x ^{2}+ y ^{2}-3 x + fy + C =0$
This will represent a circle, if
$-\frac{a}{2}=1$ $\left[\because\right.$ Coeff. of $x^{2}=$ Coeff. of $\left.y^{2}\right]$
and,$\frac{9}{4}+\frac{ f ^{2}}{4}- C >0 $
$\left[\right.$ Using $: g ^{2}+ f ^{2}- c >0$ ]
$\Rightarrow a=-2$ and $9+f^{2}-4 C>0$