Question

Find the solution of the differential equation:
$xdx+ydy+\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0$ will be
${{x}^{2}}+{{y}^{2}}-2{{\tan }^{-1}}x=c$
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)=c$
${{x}^{2}}+{{y}^{2}}+{{\tan }^{-1}}\left( \dfrac{y}{x} \right)=c$
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{x}{y} \right)=c$

Answer 1

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

Complete step by step answer:
Given differential equation is
$xdx+ydy+\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=0..............\left( i \right)$
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 $\int{xdx,\int{ydx\Rightarrow \dfrac{{{x}^{2}}}{2},\dfrac{{{y}^{2}}}{2}}}$ respectively by using the identity $\int{{{x}^{n}}=\dfrac{{{x}^{n}}+1}{n+1}}$
So, we get
$\dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+\int{\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}+c=0.............\left( ii \right)}$
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
$\dfrac{1}{1+{{x}^{2}}}\Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{n}^{2}}}$
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
${{\left( fg\left( x \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)...............\left( iii \right)$
Division rule of derivative is given as
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-\dfrac{du}{dx}}{{{v}^{2}}}................\left( iv \right)$
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
$\begin{aligned} & \dfrac{d}{dx}{{\tan }^{-1}}\left( \dfrac{y}{x} \right)=\dfrac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}\dfrac{x\dfrac{dy}{dx}-y\dfrac{dx}{dx}}{{{x}^{2}}} \\\ & \dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{x\dfrac{dy}{dx}-y}{{{x}^{2}}+{{y}^{2}}} \\\ & \Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\dfrac{1}{dx}\left( \dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}} \right) \\\ \end{aligned}$
Cancel out ‘dx’ from both sides, we get
$d\left( {{\tan }^{-1}}\dfrac{y}{x} \right)=\left( \dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}} \right)...............\left( v \right)$
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
$\dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+\int{d\left( {{\tan }^{-1}}\dfrac{y}{x} \right)+{{c}_{1}}=0}$
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
$\begin{aligned} & \dfrac{{{x}^{2}}}{2}+\dfrac{{{y}^{2}}}{2}+{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+{{c}_{1}}=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+2{{c}_{1}}=0 \\\ \end{aligned}$
Hence, solution of the given differential equation will be
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)+2{{c}_{1}}=0$
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)=-2{{c}_{1}}$
Replace $-2{{c}_{1}}$ by c, hence we get
${{x}^{2}}+{{y}^{2}}+2{{\tan }^{-1}}\left( \dfrac{y}{x} \right)=c$
So, option (c) is correct.
Note: One may try to use linear, homogenous 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,
$\int{\dfrac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=\int{d\left( {{\tan }^{-1}}\left( \dfrac{y}{d} \right) \right)}}$
Is the key point of the question. Another approach for this question would be that we can put $x=r\sin \theta ,y=r\cos \theta$ to the given expression.