Question

Solve the equation $xdx+ydy+\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0$

Answer 1

Hint: Observe the term $\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}$ which is written in form of differentiation of ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$.
   $\dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{x\dfrac{dy}{dx}-y}{{{x}^{2}}+{{y}^{2}}}$.
And use the formula of $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n}}+1}{n+1}$ to solve the given differential equation.

Complete Step-by-Step solution:
Given differential equation is
$xdx+ydy+\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0$ …………….. (i)
Here, we need to observe the term $\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}$ and use the exact differentiable approach as $\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}$ is the direct derivative of some term and we have to observe that only. So let us directly integrate the equation (i) with the help of the mentioned method (directly applying integration to the given differential equation). So, we get:
$\int{xdx+\int{ydy+}}\int{\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}}=0$
We know the integration of ${{x}^{n}}$ with respect ‘x’ is given as
 $\dfrac{{{x}^{n+1}}}{n+1}\Rightarrow \int{{{x}^{n}}dx=}\dfrac{{{x}^{n+1}}}{n+1}$ .
So, we can replace the terms $\int{xdx,\int{ydx\to \dfrac{{{x}^{2}}}{2},}}\dfrac{{{y}^{2}}}{2}$ respectively by using the identity $\int{{{x}^{n}}}=\dfrac{{{x}^{n+1}}}{n+1}$ So, we get
$\dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+\int{\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}}+c=0$ ………… (ii)
Now, let us differentiate the term ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ with respect to ‘x’, so, we need to find
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)=?$
As we know the derivative of ${{\tan }^{-1}}x$ w.r.t x is
 $\dfrac{1}{1+{{x}^{2}}}\Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}$ .
And we need to use chain rule and division rule of derivative to differentiate the expression ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$. Chain rule will be used because ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$is a composite function i.e. algebraic function with inverse trigonometric function. Chain rule is given as
${{\left( f\left( g\left( x \right) \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)$ …………… (iii)
Division rule of derivative is given as
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ ……………….. (iv)
Where u and v are two functions in fraction
Now, let us differentiate $\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)$
So, we get
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)=\dfrac{1}{1+{{\left( \dfrac{y}{x} \right)}^{2}}}\dfrac{d}{dx}\left( \dfrac{y}{x} \right)$
Where, we know $\dfrac{d}{dx}{{\tan }^{-1}}x=\dfrac{1}{1+{{x}^{2}}}$. So, we get
$\dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\dfrac{d}{dx}\left( \dfrac{y}{x} \right)$
Now, use the division rule of derivative given in equation (iv) by putting u = y and v = x.
So, we get
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)=\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\dfrac{x\dfrac{dy}{dx}-y\dfrac{dx}{dx}}{{{x}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{x\dfrac{dy}{dx}-y}{{{x}^{2}}+{{y}^{2}}} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{1}{dx}\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Cancel out ‘dx’ from both the sides, we get
$d\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}$ ………………. (v)
Now, we can replace$\left( \dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}} \right)$ in the equation (ii) by using the equation (v).
So, we get equation (ii) as
$\dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+\int{d\left( {{\tan }^{-1}}\dfrac{y}{x} \right)+c=0}$
Now, as we know integration of dx is x, similarly, integration of $d\left( {{\tan }^{-1}}\dfrac{y}{x} \right)\Rightarrow {{\tan }^{-1}}\left( \dfrac{y}{x} \right)$
So, we get
$\dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+c=0,{{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+2c=0$
Hence, solution of the given differential equation will be
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+2c=0$

Note: One may try to use linear, homogeneous or variable separable methods to solve the given differential equation, which will be very complex approaches for these kinds of questions. So, try to observe the pattern and apply the required method.
Observing $\int{\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}}=\int{d\left( {{\tan }^{-1}}\left( \dfrac{y}{x} \right) \right)}$ is the key point of the question.
Another approach for this question would be that we can put $x=r\sin \theta ,y=r\sin \theta $ to the given expression and simplify it as:
$\begin{align}
  & x=r\sin \theta \\
 & y=r\sin \theta \\
\end{align}$
Hence, we get,
$\begin{align}
  & \dfrac{dx}{d\theta }=r\cos \theta ,\dfrac{dy}{d\theta }=-rsin\theta , \\
 & dx=r\cos \theta d\theta ,dy=-rsin\theta d\theta \\
\end{align}$
Replace dx and dy by $r\cos \theta d\theta ,-rsin\theta d\theta $ respectively in the given expression. Hence, solve it further.