Question

A solution to the differential equation: $\left( \dfrac{dy}{dx} \right)-x\dfrac{dy}{dx}+y=0$ is?

Answer 1

Hint: Take $\dfrac{dy}{dx}$ common from the first two terms. Take all the terms containing ‘x’ one side and all the terms containing ‘y’ at another side. Integrate the side containing ‘x’ with respect to the variable x and the side containing ‘y’ with respect to the variable y. Add an arbitrary constant or constant of integration at one of the sides to get the answer.

Complete step-by-step solution -
We have been given the differential equation: $\left( \dfrac{dy}{dx} \right)-x\dfrac{dy}{dx}+y=0$.
Taking $\dfrac{dy}{dx}$ common from the first two terms, we get,
$\left( \dfrac{dy}{dx} \right)\left( 1-x \right)+y=0$
This can be written as,
$\begin{aligned} & y=-\dfrac{dy}{dx}\left( 1-x \right) \\\ & \Rightarrow y=\dfrac{dy}{dx}\left( x-1 \right) \\\ \end{aligned}$
Now, separating the terms containing ‘y’ to the L.H.S and the terms containing ‘x’ to the R.H.S, we get,
$\dfrac{dy}{y}=\dfrac{dx}{\left( x-1 \right)}$
Integrating the above expression, we get,
$\int{\dfrac{dy}{y}=}\int{\dfrac{dx}{\left( x-1 \right)}}$
We know that, $\int{\dfrac{dx}{x}=}\log x$. Therefore,
$\log y=\log \left( x-1 \right)+\log c$, where $\log c$ is any arbitrary constant.
Now, using the property of logarithm given by: $\log m+\log n=\log mn$, we get,
$\log y=\log c\left( x-1 \right)$
Removing log from both sides, we get,
$\begin{aligned} & y=c\left( x-1 \right) \\\ & \Rightarrow \dfrac{y}{\left( x-1 \right)}=c \\\ \end{aligned}$

Note: One may note that, we have chosen ‘$\log c$’ and not ‘c’ as our arbitrary constant or you can say constant of integration. The reason is that we can easily use the properties of logarithm and simplify the expression. If we will use ‘c’ as the arbitrary constant then we have to use the property of exponents to simplify which will be lengthy. However, the answer will not change. The above method of finding a solution is known as a variable separable method.