Question

Solve the differential equation : $\left(x-\sqrt{xy}\right)dy=ydx$

Answer 1

Consider the given equation.

$(x-\sqrt{xy})dy=ydx$

$\frac{dy}{dx}=\frac{y}{(x-\sqrt{xy})}$ …… (1)

Put $y=vx$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Therefore,

$\frac{vx}{(x-\sqrt{x^{2}v})}=v+x\frac{dv}{dx}$

$\frac{v}{(1-\sqrt{v})}=v+x\frac{dv}{dx}$

$x\frac{dv}{dx}=\frac{v}{(1-\sqrt{v})}-v$

$x\frac{dv}{dx}=\frac{v^{3/2}}{(1-\sqrt{v})}$

$\left(\frac{1-\sqrt{v}}{v^{3/2}}\right)dv=\frac{dx}{x}$

On taking integral both sides, we get

$\int \left(\frac{1-\sqrt{v}}{v^{3/2}}\right)dv=\int \frac{dx}{x}$

$\int (\frac{1}{v^{3/2}}-\frac{1}{v})dv=\int \frac{dx}{x}$

$\frac{-2}{\sqrt{v}}-\ln \left(v\right)=\ln x+C$

On putting the value of $v$, we get

$\frac{-2}{\sqrt{\frac{y}{x}}}-\ln \left(\frac{y}{x}\right)=\ln x+C$

$-2\sqrt{\frac{x}{y}}-\ln \left(\frac{y}{x}\right)=\ln x+C$

$-\ln x-\ln \left(\frac{y}{x}\right)=2\sqrt{\frac{x}{y}}+C$

$-\ln \left(y\right)=2\sqrt{\frac{x}{y}}+C$

Hence, this is the answer.Consider the given equation.

$(x-\sqrt{xy})dy=ydx$

$\frac{dy}{dx}=\frac{y}{(x-\sqrt{xy})}$ …… (1)

Put $y=vx$

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Therefore,

$\frac{vx}{(x-\sqrt{x^{2}v})}=v+x\frac{dv}{dx}$

$\frac{v}{(1-\sqrt{v})}=v+x\frac{dv}{dx}$

$x\frac{dv}{dx}=\frac{v}{(1-\sqrt{v})}-v$

$x\frac{dv}{dx}=\frac{v^{3/2}}{(1-\sqrt{v})}$

$\left(\frac{1-\sqrt{v}}{v^{3/2}}\right)dv=\frac{dx}{x}$

On taking integral both sides, we get

$\int \left(\frac{1-\sqrt{v}}{v^{3/2}}\right)dv=\int \frac{dx}{x}$

$\int (\frac{1}{v^{3/2}}-\frac{1}{v})dv=\int \frac{dx}{x}$

$\frac{-2}{\sqrt{v}}-\ln \left(v\right)=\ln x+C$

On putting the value of $v$, we get

$\frac{-2}{\sqrt{\frac{y}{x}}}-\ln \left(\frac{y}{x}\right)=\ln x+C$

$-2\sqrt{\frac{x}{y}}-\ln \left(\frac{y}{x}\right)=\ln x+C$

$-\ln x-\ln \left(\frac{y}{x}\right)=2\sqrt{\frac{x}{y}}+C$

$-\ln \left(y\right)=2\sqrt{\frac{x}{y}}+C$

Hence, this is the answer.