Question

If solution of the differential equation $y^{\prime }+\frac{y}{x}-\sqrt{y}=0$ $y(1)=0$ is $f(x)$, then $[9f(9)]$ equals (where $[.]$ greatest integer function)

• A. 70
• B. 0
• C. 75
• D. 61

Answer 1

Answer: (C)

$y^{\prime }+\frac{y}{x}=y^{\frac{1}{2}}$
$\Rightarrow y^{-\frac{1}{2}}{y}^{\prime }+\frac{1}{x}{y}^{\frac{1}{2}}=1$
Let $y^{\frac{1}{2}}=v$
$\Rightarrow 2v^{\prime }+\frac{1}{x}v=1$
$\Rightarrow v^{\prime }+\frac{1}{2x}v=\frac{1}{2}$
Here $P=\frac{1}{2x},Q=\frac{1}{2}$
$IF=e^{\int pdx}=e^{\int \frac{1}{2x}dx}=e^{\frac{1}{2}In|x|}=|x{|}^{\frac{1}{2}}$
So, $v\times IF=\int IF\times Qdx+c$
$v\times |x{|}^{\frac{1}{2}}=\int |x{|}^{\frac{1}{2}}\times \frac{1}{2}dx+c$
$\Rightarrow y^{\frac{1}{2}}|x{|}^{\frac{1}{2}}=\frac{2|x{|}^{\frac{3}{2}}}{3}\times \frac{1}{2}+c$
$y^{\frac{1}{2}}=\frac{1}{3}x+c{x}^{-\frac{1}{2}}$
Now, $y(1)=0\Rightarrow 0=\frac{1}{3}+c$
$\Rightarrow c=-\frac{1}{3}$
$\therefore y={\left(\frac{1}{3}x-\frac{1}{3}{x}^{-\frac{1}{2}}\right)}^2$
$\Rightarrow y=\frac{x^3-2x^{\frac{3}{2}}+1}{9x}$
$\Rightarrow y=\frac{1}{9}\left(x^2-2x^{\frac{1}{2}}+\frac{1}{x}\right)$
$\therefore [9f(9)]=75$