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Q.
If solution of the differential equation $y^{\prime}+\frac{y}{x}-\sqrt{y}=0$ $y(1)=0$ is $f(x)$, then $[9 f(9)]$ equals (where $[.]$ greatest integer function)
Differential Equations
Solution:
$y'+\frac{y}{x}=y^{\frac{1}{2}}$
$\Rightarrow y^{-\frac{1}{2}} y'+\frac{1}{x} y^{\frac{1}{2}}=1$
Let $y^{\frac{1}{2}} = v$
$\Rightarrow 2 v'+\frac{1}{x} v=1 $
$\Rightarrow v'+\frac{1}{2 x} v=\frac{1}{2}$
Here $P=\frac{1}{2 x}, Q=\frac{1}{2}$
$IF =e^{\int p d x}=e^{\int \frac{1}{2 x} d x}=e^{\frac{1}{2} In|x|}=|x|^{\frac{1}{2}}$
So, $v \times IF =\int I F \times Q d x+c$
$v \times|x|^{\frac{1}{2}}=\int|x|^{\frac{1}{2}} \times \frac{1}{2} d x+c$
$ \Rightarrow y^{\frac{1}{2}}|x|^{\frac{1}{2}}=\frac{2|x|^{\frac{3}{2}}}{3} \times \frac{1}{2}+c$
$y^{\frac{1}{2}}=\frac{1}{3} x+c x^{-\frac{1}{2}}$
Now, $y(1)=0 \Rightarrow 0=\frac{1}{3}+c$
$ \Rightarrow c=-\frac{1}{3}$
$\therefore y=\left(\frac{1}{3} x-\frac{1}{3} x^{-\frac{1}{2}}\right)^{2} $
$\Rightarrow y=\frac{x^{3}-2 x^{\frac{3}{2}}+1}{9 x}$
$\Rightarrow y=\frac{1}{9}\left(x^{2}-2 x^{\frac{1}{2}}+\frac{1}{x}\right)$
$\therefore [9 f(9)]=75$