Question

Find the solution of $\left( x\sqrt{xy}\left( x+y \right)-y \right)dx+\left( y\sqrt{xy}\left( x+y \right)-x \right)dy=0$.
A. ${{x}^{2}}+{{y}^{2}}=2{{\tan }^{-1}}\sqrt{\dfrac{y}{x}}+c$
B. ${{x}^{2}}+{{y}^{2}}=4{{\tan }^{-1}}\sqrt{\dfrac{y}{x}}+c$
C. ${{x}^{2}}+{{y}^{2}}={{\tan }^{-1}}\sqrt{\dfrac{y}{x}}+c$
D. none of these

Answer 1

Hint: We have to find the differential forms of many consecutive composite functions. We keep rearranging the equation in such a way that the composite functions keep forming the differential forms. At the end we integrate the whole differential to find the solution.

Complete step by step answer:
We take the given differential form and get $\left( x\sqrt{xy}\left( x+y \right)-y \right)dx+\left( y\sqrt{xy}\left( x+y \right)-x \right)dy=0$.
We try to form the general differential forms where
\[\begin{align}
  & \left( x\sqrt{xy}\left( x+y \right)-y \right)dx+\left( y\sqrt{xy}\left( x+y \right)-x \right)dy=0 \\
 & \Rightarrow \sqrt{xy}\left( x+y \right)\left[ xdx+ydy \right]=xdy+ydx \\
 & \Rightarrow 2\sqrt{xy}\left( x+y \right)\left[ xdx+ydy \right]=2\left( xdy+ydx \right) \\
\end{align}\]
We try to find the differential form of ${{x}^{2}}+{{y}^{2}}$.
So, $d\left( {{x}^{2}}+{{y}^{2}} \right)=d\left( {{x}^{2}} \right)+d\left( {{y}^{2}} \right)=2xdx+2ydy$.
We replace the values and get
\[\begin{align}
  & 2\sqrt{xy}\left( x+y \right)\left[ xdx+ydy \right]=2\left( xdy+ydx \right) \\
 & \Rightarrow 2xdx+2ydy=\,\dfrac{2\left( xdy+ydx \right)}{\sqrt{xy}\left( x+y \right)} \\
 & \Rightarrow d\left( {{x}^{2}}+{{y}^{2}} \right)=\,\dfrac{2\left( xdy+ydx \right)}{\sqrt{xy}\left( x+y \right)} \\
\end{align}\]
We now try to find the differential form of $\dfrac{y}{x}$.
So, $d\left( \dfrac{y}{x} \right)=\dfrac{xdy+ydx}{{{x}^{2}}}$.
We rearrange the equation \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,\dfrac{2\left( xdy+ydx \right)}{\sqrt{xy}\left( x+y \right)}\] and get
\[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,2\dfrac{\left( xdy+ydx \right)}{{{x}^{2}}}\times \dfrac{x}{\left( x+y \right)}\times \dfrac{x}{\sqrt{xy}}\]
Replacing the value, we get \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,2d\left( \dfrac{y}{x} \right)\times \dfrac{x}{\left( x+y \right)}\times \dfrac{x}{\sqrt{xy}}\].
We now try to find the differential form of $\sqrt{\dfrac{y}{x}}$.
So, $d\left( \sqrt{\dfrac{y}{x}} \right)=\dfrac{1}{2\sqrt{\dfrac{y}{x}}}\times d\left( \dfrac{y}{x} \right)=\dfrac{x}{2\sqrt{xy}}\times d\left( \dfrac{y}{x} \right)$.
We rearrange the equation \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,2d\left( \dfrac{y}{x} \right)\times \dfrac{x}{\left( x+y \right)}\times \dfrac{x}{\sqrt{xy}}\] and get
\[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,4\dfrac{x}{\left( x+y \right)}\times \dfrac{x}{2\sqrt{xy}}d\left( \dfrac{y}{x} \right)\]
Replacing the value, we get \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,4\dfrac{x}{\left( x+y \right)}\times d\left( \sqrt{\dfrac{y}{x}} \right)\]
We now try to find the differential form of ${{\tan }^{-1}}\sqrt{\dfrac{y}{x}}$.
So, $d\left( {{\tan }^{-1}}\sqrt{\dfrac{y}{x}} \right)=\dfrac{1}{1+{{\left( \sqrt{\dfrac{y}{x}} \right)}^{2}}}d\left( \sqrt{\dfrac{y}{x}} \right)=\dfrac{1}{1+\dfrac{y}{x}}d\left( \sqrt{\dfrac{y}{x}} \right)=\dfrac{x}{x+y}d\left( \sqrt{\dfrac{y}{x}} \right)$.
Replacing the value in \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,4\dfrac{x}{\left( x+y \right)}\times d\left( \sqrt{\dfrac{y}{x}} \right)\], we get \[d\left( {{x}^{2}}+{{y}^{2}} \right)=\,4\dfrac{x}{\left( x+y \right)}\times d\left( \sqrt{\dfrac{y}{x}} \right)=4d\left( {{\tan }^{-1}}\sqrt{\dfrac{y}{x}} \right)\]
The total differential form is \[d\left( {{x}^{2}}+{{y}^{2}} \right)=4d\left( {{\tan }^{-1}}\sqrt{\dfrac{y}{x}} \right)\].
We take indefinite integral both sides to get
\[\begin{align}
  & \int{d\left( {{x}^{2}}+{{y}^{2}} \right)}=4\int{d\left( {{\tan }^{-1}}\sqrt{\dfrac{y}{x}} \right)} \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}=4{{\tan }^{-1}}\sqrt{\dfrac{y}{x}}+c \\
\end{align}\]

The correct option is B.

Note: We can also assume the variable where $\dfrac{y}{x}={{v}^{2}}$, but in that case we have to take the new variable. The solution is the same but the process becomes more complicated. That's why it is better to use differential form than using different variables.