Question

What is a solution to the differential equation $\dfrac{dy}{dx}=\dfrac{1}{x}$ ?

Answer 1

Hint : 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 $\dfrac{dy}{dx}=\dfrac{1}{x}$ . We take one constant term in the form of logarithm for the integration. We get the equation of a circle.

Complete step by step solution:
We have given a differential equation $\dfrac{dy}{dx}=\dfrac{1}{x}$ .
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 v to form the differential form.
So, $\dfrac{dy}{dx}=\dfrac{1}{x}\Rightarrow dy=\dfrac{dx}{x}$
We now need to integrate the function $dy=\dfrac{dx}{x}$ to find the solution of the differential equation. We get $\int{dy}=\int{\dfrac{dx}{x}}+c$ .
We know the integral form of $\int{\dfrac{dx}{x}}=\log \left| x \right|$ .
Simplifying the differential form, we get
$\begin{aligned} & \int{dy}=\int{\dfrac{dx}{x}}+c \\\ & \Rightarrow y=\log \left| x \right|+\log \left| k \right| \\\ & \Rightarrow 2y=2\log \left| kx \right|=\log \left( {{k}^{2}}{{x}^{2}} \right) \\\ \end{aligned}$
Here $c$ is another constant. We simplify to get
$\begin{aligned} & 2y=\log \left( {{k}^{2}}{{x}^{2}} \right) \\\ & \Rightarrow K{{x}^{2}}={{e}^{2y}} \\\ \end{aligned}$
The solution of the differential equation $\dfrac{dy}{dx}=\dfrac{1}{x}$ is $K{{x}^{2}}={{e}^{2y}}$ . $K$ is also constant.
So, the correct answer is “ $K{{x}^{2}}={{e}^{2y}}$ ”.

Note : The solution of the differential equation is the equation of $K{{x}^{2}}={{e}^{2y}}$ . The first order differentiation of $K{{x}^{2}}={{e}^{2y}}$ gives the tangent of the circle for a certain point which is equal to $\dfrac{dy}{dx}=\dfrac{1}{x}$ .