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{align}
  & \int{\dfrac{dy}{y}}=\int{\dfrac{dx}{x}}+k \\
 & \Rightarrow \log \left| y \right|=\log \left| x \right|+\log \left| c \right| \\
\end{align} $
Here $ c $ is another constant where $ k=\log \left| c \right| $ .
We now take all logarithms in one side and get
 $ \begin{align}
  & \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{align} $
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 $ .