Question

The solution of the differential equation $xdy-ydx=\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)dx$ is
(A). $y-\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{x}^{2}}$
(B). $y+\sqrt{{{x}^{2}}+{{y}^{2}}}=C{{x}^{2}}$
(C). $y+\sqrt{{{x}^{2}}+{{y}^{2}}}+C{{x}^{2}}=0$
(D). None of these

Answer 1

Hint: Any differential equation would be of homogeneous type if the given function F (x, y) follows the equation $F\left( \lambda x,\lambda y \right)={{\lambda }^{n}}F\left( x,y \right)$ where n is an integer. And we can solve any differential equation in variables ‘x’ and ‘y’ by putting y = vx or x = vy where ‘v’ is another variable. Now, try to solve it. And use the integration of $\dfrac{1}{{{x}^{2}}+{{a}^{2}}}$given as $\int{\dfrac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\log \left( x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right)}$.

Complete step-by-step solution -
Given differential equation is,
$xdy-ydx=\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)dx.............\left( i \right)$
On dividing the whole equation (i) by ‘dx’ we get,
$\begin{align}
  & x\dfrac{dy}{dx}-y=\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right) \\
 & or \\
 & \dfrac{dy}{dx}=\dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}+y}{x}...........\left( ii \right) \\
\end{align}$
Now, we need to observe the type of differential equation i.e. variable separable or linear or homogeneous equation.
Here, we can observe that we cannot separate the variables ‘x’ and ‘y’ and cannot form a linear relation as well.
So, let us check whether the differential equation is homogeneous or not?
Let $F\left( x,y \right)=\dfrac{y+\sqrt{{{x}^{2}}+{{y}^{2}}}}{x}$
Replace $\left( x,y \right)\ by\ \left( \lambda x,\lambda y \right)$ and try to get an equation of type ${{\lambda }^{n}}F\left( x,y \right)$.
Hence,
$\begin{align}
  & F\left( \lambda x,\lambda y \right)=\dfrac{\lambda y+\sqrt{{{\left( \lambda x \right)}^{2}}+{{\left( \lambda y \right)}^{2}}}}{\lambda x} \\
 & F\left( \lambda x,\lambda y \right)=\dfrac{\lambda y+\lambda \sqrt{{{x}^{2}}+{{y}^{2}}}}{\lambda x} \\
 & F\left( \lambda x,\lambda y \right)=\lambda {}^\circ F\left( x,y \right) \\
\end{align}$
Hence, the given differential equation is of homogeneous type.
So, take y = vx in equation (ii).
Now, differentiating y = vx, we get;
$\dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx}$
Where, we used relation
$\dfrac{d}{dx}\left( u.v \right)=u\dfrac{du}{dx}+v\dfrac{du}{dx}$
Hence, we get
$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}..............\left( iii \right)$
So, equation (ii) can be written as
$\begin{align}
  & v+x\dfrac{dv}{dx}=\dfrac{\sqrt{{{x}^{2}}+{{\left( vx \right)}^{2}}}+vx}{x} \\
 & v+x\dfrac{dv}{dx}=\dfrac{x\sqrt{1+{{v}^{2}}}+vx}{x} \\
 & v+x\dfrac{dv}{dx}=\sqrt{1+{{v}^{2}}}+v \\
 & x\dfrac{dv}{dx}=\sqrt{1+{{v}^{2}}}..............\left( iv \right) \\
\end{align}$
Now, it’s a simple variable separable differential equation and can be written as,
$\int{\dfrac{dv}{\sqrt{1+{{v}^{2}}}}=\int{\dfrac{1}{x}dx...........\left( v \right)}}$
Now, we know,
$\begin{align}
  & \int{\dfrac{1}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx={{\log }_{e}}\left( x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right)} \\
 & \int{\dfrac{1}{x}dx}={{\log }_{e}}x \\
\end{align}$
Hence, equation (v) can be written as,
${{\log }_{e}}\left( 1+\sqrt{1+{{v}^{2}}} \right)={{\log }_{e}}x+{{\log }_{e}}c$
Where ${{\log }_{e}}c$ is a constant.
Now, we can use property of logarithm as,
${{\log }_{a}}m+{{\log }_{a}}n={{\log }_{a}}\left( mn \right)$
So, we get
${{\log }_{e}}\left( 1+\sqrt{1+{{v}^{2}}} \right)={{\log }_{e}}xc$
And hence,
$1+\sqrt{1+{{v}^{2}}}=cx$
Now, putting the value of $v=\dfrac{y}{x}$ by relation y = vx. Hence, above equation can be written as’
$\begin{align}
  & 1+\sqrt{1+{{\left( \dfrac{y}{x} \right)}^{2}}}=cx \\
 & 1+\dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{x}=cx \\
 & or \\
 & x+\sqrt{{{x}^{2}}+{{y}^{2}}}=c{{x}^{2}} \\
\end{align}$
And hence, option (B) is the correct answer.

Note: Calculation is the important side of the question.
Observation is the key of the differential equations. One can use x = vy relation as well for solving the given equation.
One may get confused why constant term is taken in log in solution.
Reason is simple, for the simplicity of the equation so that we can remove ‘log’ from the solution part. And we can use ${{\log }_{e}}c\ or\ C\ or\ {{\tan }^{-1}}c$ etc. for writing the constant part in indefinite integral.