Solve the Differential Equation (x^2+1)y’=xy
In this tutorial we shall solve a differential equation of the form $$\left( {{x^2} + 1} \right)y’ = xy$$ by using the separating the variables method.
The differential equation of the form is given as
\[\left( {{x^2} + 1} \right)y’ = xy\]
This differential equation can also be written as
\[\left( {{x^2} + 1} \right)\frac{{dy}}{{dx}} = xy\]
Separating the variables, the given differential equation can be written as
\[\frac{1}{y}dy = \frac{x}{{{x^2} + 1}}dx\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]
Keep in mind that in the separating variable technique the terms $$dy$$ and $$dx$$ are placed in the numerator with their respective variables.
Now integrating both sides of the equation (i), we have
\[\begin{gathered} \int {\frac{1}{y}dy = \int {\frac{x}{{{x^2} + 1}}dx} } \\ \Rightarrow \int {\frac{1}{y}dy = \frac{1}{2}\int {\frac{{2x}}{{{x^2} + 1}}dx} } \\ \end{gathered} \]
Using the formulas of integration $$\int {\frac{{f’\left( x \right)}}{{f\left( x \right)}}dx = \ln f\left( x \right)} $$ and $$\int {\frac{1}{x}dx = \ln x} $$, we get
\[\begin{gathered} \ln y = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \ln c \\ \Rightarrow \ln y = \ln {\left( {{x^2} + 1} \right)^{\frac{1}{2}}} + \ln c \\ \Rightarrow \ln y = \ln \sqrt {{x^2} + 1} + \ln c \\ \Rightarrow \ln y = \ln c\sqrt {{x^2} + 1} \\ \Rightarrow y = c\sqrt {{x^2} + 1} \\ \end{gathered} \]
This is the required solution of the given differential equation.