Question

Solve the following differential equation.
$2{{x}^{2}}\dfrac{dy}{dx}-2xy+{{y}^{2}}=0$ .

Answer 1

Hint: In order to solve this problem, we need to solve using the proper substitution. We need to use the substitution of $y=vx$ . Also, we must know the product rule of differentiation which is given $\dfrac{du\left( x \right)v\left( x \right)}{dx}=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+\dfrac{du\left( x \right)}{dx}v\left( x \right)$ . We also need to know some of the standard integration formulas such as $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ and $\int{\dfrac{1}{x}=\ln \left( x \right)}$ .

Complete step by step answer:
We are given a differential equation to solve.
As we can see that it is not possible to solve this equation by separating the y and x.
So, in order to separate the variable, we need to use substitution.
The equation is $2{{x}^{2}}\dfrac{dy}{dx}-2xy+{{y}^{2}}=0$ .
Let’s use the substitution, $y=vx$ .
Differentiating both sides with respect to x, we get,
$\dfrac{dy}{dx}=v\dfrac{dx}{dx}+\dfrac{dv}{dx}.x$
Solving this we get,
$\dfrac{dy}{dx}=v+\dfrac{dv}{dx}.x$
Substituting this value in the differential equation we get,
$\begin{align}
  & 2{{x}^{2}}\dfrac{dy}{dx}-2xy+{{y}^{2}}=0 \\
 & \dfrac{dy}{dx}=\dfrac{2xy-{{y}^{2}}}{2{{x}^{2}}} \\
 & \dfrac{dy}{dx}=\dfrac{2\left( \dfrac{y}{x} \right)-{{\left( \dfrac{y}{x} \right)}^{2}}}{2} \\
 & v+\dfrac{dv}{dx}.x=\dfrac{2v-{{v}^{2}}}{2} \\
\end{align}$
Solving this equation, we get,
$\begin{align}
  & v+\dfrac{dv}{dx}.x=\dfrac{2v-{{v}^{2}}}{2} \\
 & v+\dfrac{dv}{dx}.x=v-\dfrac{{{v}^{2}}}{2} \\
 & x\dfrac{dv}{dx}=\dfrac{-{{v}^{2}}}{2} \\
\end{align}$
Separating the variables we get,
$\begin{align}
  & x\dfrac{dv}{dx}=\dfrac{-{{v}^{2}}}{2} \\
 & \dfrac{2}{{{v}^{2}}}dv=-\dfrac{dx}{x} \\
\end{align}$
Taking the integration on both sides with respect to respective variables, we get,
$\int{\dfrac{2}{{{v}^{2}}}dv}=\int{-\dfrac{dx}{x}}$
Using the formula of integration,
$\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ and $\int{\dfrac{1}{x}=\ln \left( x \right)}$, we get,
$\begin{align}
  & \int{\dfrac{2}{{{v}^{2}}}dv}=\int{-\dfrac{dx}{x}} \\
 & 2\dfrac{{{v}^{-2+1}}}{-2+1}=-\ln x+\ln C \\
\end{align}$
Where $\ln C$ is the constant of integration.
Solving this we get,
$\begin{align}
  & 2\dfrac{{{v}^{-1}}}{-1}=-\ln x+\ln C \\
 & \dfrac{-2}{v}=-\ln x+\ln C \\
 & -\dfrac{2}{v}=-\ln \left( xC \right) \\
 & {{e}^{\left( \dfrac{2}{v} \right)}}=xC \\
\end{align}$
Resubstituting the value of v we get,
${{e}^{\dfrac{2x}{y}}}=xC$

Hence the solution of the differential equation is ${{e}^{\dfrac{2x}{y}}}=xC$ .

Note: In this problem, while performing the differentiation of the substitution we need to use the product rule of differentiation. Also, as we can see that we are not considering the sign for the value of constant as all the signs can be included in the constant itself because any sign multiplied by a constant is still a constant.