Question

What is the solution of the differential equation $\dfrac{dy}{dx}=\dfrac{2x}{y}$?

Answer 1

We first explain the term $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation once to find all the solutions of the differential equation $\dfrac{dy}{dx}=\dfrac{2x}{y}$. We take one constant term in the form of logarithm for the integration. We get the equation of a circle.

Complete step by step solution:
We have given a differential equation $\dfrac{dy}{dx}=\dfrac{2x}{y}$.
Here $\dfrac{dy}{dx}$ defines the first order differentiation which is expressed as $\dfrac{dy}{dx}=\dfrac{d}{dx}\left( y \right)$.
The main function is $y=f\left( x \right)$.
We have to find the antiderivative or the integral form of the equation.
We first interchange the terms in $\dfrac{dy}{dx}=\dfrac{2x}{y}$ to form the differential form.
So, $\dfrac{dy}{dx}=\dfrac{2x}{y}\Rightarrow ydy=2xdx$
We now need to integrate the function $ydy=2xdx$ to find the solution of the differential equation. We get $\int{ydy}=\int{2xdx}$.
We know the integral form of $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$.
Simplifying the differential form, we get
$\begin{aligned} & \int{ydy}=\int{2xdx} \\\ & \Rightarrow \dfrac{{{y}^{2}}}{2}=2\times \dfrac{{{x}^{2}}}{2}+k \\\ & \Rightarrow {{y}^{2}}=2{{x}^{2}}+c \\\ \end{aligned}$
Here $c$ is another constant where $c=2k$.
The equation becomes ${{y}^{2}}=2{{x}^{2}}+c$.
The solution of the differential equation $\dfrac{dy}{dx}=\dfrac{2x}{y}$ is ${{y}^{2}}=2{{x}^{2}}+c$.

Note: The solution of the differential equation is the equation of a circle. The first order differentiation of ${{y}^{2}}=2{{x}^{2}}+c$ gives the tangent of the circle for a certain point which is equal to $\dfrac{dy}{dx}=\dfrac{2x}{y}$.