Question

What is the solution of the differential equation $x{{y}^{'}}=y$ ?

Answer 1

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

Complete step-by-step answer :
We know that ${{y}^{'}}=\dfrac{dy}{dx}$ which converts the equation $x{{y}^{'}}=y$ into $x\dfrac{dy}{dx}=y$ .
We have given a differential equation $\dfrac{dy}{dx}=\dfrac{y}{x}$ .
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 $\dfrac{dy}{dx}=\dfrac{y}{x}$ to form the differential form.
So, $\dfrac{dy}{dx}=\dfrac{y}{x}\Rightarrow \dfrac{dy}{y}=\dfrac{dx}{x}$
We now need to integrate the function $\dfrac{dy}{y}=\dfrac{dx}{x}$ to find the solution of the differential equation. We get $\int{\dfrac{dy}{y}}=\int{\dfrac{dx}{x}}+k$ .
We know the integral form of $\int{\dfrac{dx}{x}}=\log \left| x \right|$ .
Simplifying the differential form, we get
$\begin{aligned} & \int{\dfrac{dy}{y}}=\int{\dfrac{dx}{x}}+k \\\ & \Rightarrow \log \left| y \right|=\log \left| x \right|+\log \left| c \right| \\\ \end{aligned}$
Here $c$ is another constant where $k=\log \left| c \right|$ .
We now take all logarithms in one side and get
$\begin{aligned} & \log \left| y \right|-\log \left| x \right|=\log \left| c \right| \\\ & \Rightarrow \left| \dfrac{y}{x} \right|=\left| c \right| \\\ & \Rightarrow \dfrac{{{y}^{2}}}{{{x}^{2}}}={{c}^{2}} \\\ \end{aligned}$
Simplifying and taking ${{c}^{2}}=C$ we get ${{y}^{2}}=C{{x}^{2}}$
The solution of the differential equation $x{{y}^{'}}=y$ is ${{y}^{2}}=C{{x}^{2}}$ .
So, the correct answer is “ ${{y}^{2}}=C{{x}^{2}}$ ”.

Note : The solution of the differential equation is the equation of two straight lines. The first order differentiation of ${{y}^{2}}=C{{x}^{2}}$ gives the tangent of the circle for a certain point which is equal to $x{{y}^{'}}=y$ .