Question

Solve the following:
$\left( {{x}^{2}}+{{y}^{2}} \right)dx+\left( 2xy \right)dy=0$

Answer 1

Hint: We will be using the concepts of differential equations to solve the problem. We will first test the given differential equation for homogeneity then we will use the method of solving homogeneous differential equation to solve the problem in which we make the substitution $y=\upsilon x$ to convert the differential equation in one variable v.

Complete step-by-step answer:
Now, we have been given that,
$\begin{align}
  & \left( {{x}^{2}}+{{y}^{2}} \right)dx+\left( 2xy \right)dy=0 \\
 & \dfrac{{{x}^{2}}+{{y}^{2}}}{-2xy}=\dfrac{dy}{dx} \\
\end{align}$
Now, we will check the given differential for homogeneity for that we put $F\left( x,y \right)=\dfrac{dy}{dx}$ and will find $F\left( \lambda x,\lambda y \right)$.
$F\left( x,y \right)=\dfrac{-\left( {{x}^{2}}+{{y}^{2}} \right)}{2xy}$
So, we have,
 \[\begin{align}
  & F\left( \lambda x,\lambda y \right)=\dfrac{-\left( {{\lambda }^{2}}\left( {{x}^{2}} \right)+{{\lambda }^{2}}\left( {{y}^{2}} \right) \right.}{2{{\lambda }^{2}}xy} \\
 & =\dfrac{{{\lambda }^{2}}\left( -\left( {{x}^{2}}+{{y}^{2}} \right) \right.}{2{{\lambda }^{2}}xy} \\
 & =\dfrac{-\left( {{x}^{2}}+{{y}^{2}} \right)}{2xy} \\
 & ={{\lambda }^{0}}\dfrac{-\left( {{x}^{2}}+{{y}^{2}} \right)}{2xy} \\
 & F\left( \lambda x,\lambda y \right)={{\lambda }^{0}}F\left( x,y \right) \\
\end{align}\]
Hence, F (x, y) is a homogeneous function of degree zero. So, $\dfrac{dy}{dx}$ is a homogeneous differential equation. Therefore, we will solve it by putting $y=\upsilon x$.
Now, we differentiate it with the respect to x.
$\dfrac{dy}{dx}=\upsilon +x\dfrac{dv}{dx}$
Now, we will substitute this in,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{-\left( {{x}^{2}}+{{y}^{2}} \right)}{2xy} \\
 & \upsilon +x\dfrac{d\upsilon }{dx}=\dfrac{-\left( {{x}^{2}}+{{\upsilon }^{2}}{{x}^{2}} \right)}{2\upsilon {{x}^{2}}} \\
 & x\dfrac{d\upsilon }{dx}+\upsilon =\dfrac{-\left( 1+{{\upsilon }^{2}} \right)}{2\upsilon } \\
 & x\dfrac{d\upsilon }{dx}=\dfrac{-\left( 1+{{\upsilon }^{2}} \right)}{2\upsilon }-\upsilon \\
 & x\dfrac{d\upsilon }{dx}=\dfrac{-\left( 1+{{\upsilon }^{2}} \right)-2{{\upsilon }^{2}}}{2\upsilon } \\
 & x\dfrac{d\upsilon }{dx}=\dfrac{-1-{{\upsilon }^{2}}-2{{\upsilon }^{2}}}{2\upsilon } \\
 & x\dfrac{d\upsilon }{dx}=\dfrac{-1-3{{\upsilon }^{2}}}{2\upsilon } \\
\end{align}$
Now, on separating variable we have,
$\dfrac{dx}{x}=\dfrac{-2\upsilon }{1+3{{\upsilon }^{2}}}d\upsilon $
On integrating both sides we have,
$\int{\dfrac{dx}{x}=\int{\dfrac{-2\upsilon }{1+3{{\upsilon }^{2}}}d\upsilon }}$
Now, we know that,
$\int{\dfrac{dx}{x}=\ln \left| x \right|}$
So, we have,
$\ln \left| x \right|=\int{\dfrac{-2\upsilon }{1+3{{\upsilon }^{2}}}d\upsilon }$
Now, we take $1+3{{\upsilon }^{2}}=z$. So,
$\begin{align}
  & 6\upsilon d\upsilon =dz \\
 & 2\upsilon d\upsilon =\dfrac{dz}{3} \\
\end{align}$
On substituting this we have,
$\begin{align}
  & \ln \left| x \right|=\int{\dfrac{-dz}{3z}} \\
 & =\dfrac{-1}{3}\int{\dfrac{dz}{z}} \\
 & \ln \left| x \right|=\dfrac{-1}{3}\ln \left| z \right| \\
\end{align}$
Now, we substitute $z=1+3{{x}^{2}}\ and\ \upsilon =\dfrac{y}{x}$.
\[\begin{align}
  & \ln \left| x \right|=\dfrac{-1}{3}\ln \left| 1+3{{\upsilon }^{2}} \right|+\ln c \\
 & \ln \left| x \right|=\dfrac{-1}{3}\ln \left| 1+\dfrac{3{{y}^{2}}}{{{x}^{2}}} \right|+\ln c \\
 & \ln \left| x \right|=\dfrac{-1}{3}\ln \left| 1+\dfrac{{{x}^{2}}+3{{y}^{2}}}{{{x}^{2}}} \right|+\ln c \\
 & 3\ln \left| x \right|+\ln \left| \dfrac{{{x}^{2}}+3{{y}^{2}}}{{{x}^{2}}} \right|=3\ln c \\
 & \ln \left| {{x}^{3}}\left( \dfrac{{{x}^{2}}+3{{y}^{2}}}{{{x}^{2}}} \right) \right|=3\ln c \\
\end{align}\]
Now, let the constant $3\ln c=\ln {{c}_{1}}$.
$\begin{align}
  & \ln \left| {{x}^{3}}+3x{{y}^{2}} \right|=\ln {{c}_{1}} \\
 & {{x}^{3}}+3x{{y}^{2}}={{c}_{1}} \\
\end{align}$

Note: To solve these types of questions one should first check the differential equation for homogeneity if the equation is homogeneous then solve it using the same method otherwise use other methods to solve the problem also it has to be noted that we used the substitution that $y=\upsilon x$ to convert the whole differential equation into one variable.In last step of solution, student should remember the identities of logarithm like,
 $\log {{b}^{m}}=m\log b$ , $\log \dfrac{a}{b}=\log a-\log b$ to simplify further.