Question

How do you find the general solution to $\left( {{x^2} + 1} \right)y' = xy$?

Answer 1

Since this question is in geometric form, first try to find out if the given equation is similar to any of the trigonometric identities. If it is similar try to reconstruct the given equation into the suitable identity which will enable us to further solve the question and reach our desired solution.

Formula used:
(1) $\int {\dfrac{{x'}}{x}dx} = \left( {\ln x} \right)$
(2) $\ln xy = \ln x + \ln y$
(3) $\ln {x^2} = 2\ln x$ $\ln {x^2} = 2\ln x$

Complete step by step solution:
The above equation given to us is in two variables.
Let us separate these variables such that one variable is on one side of the equation and the other is on different sides.
$\left( {{x^2} + 1} \right)y' = xy \\\ \Rightarrow \dfrac{1}{y} \times y' = \dfrac{x}{{\left( {{x^2} + 1} \right)}} - - - (1) \\\$
Where $y'$ is nothing but $\dfrac{{dy}}{{dx}}$.
Since the above equation has been separated let us proceed further.
Let us integrate equation (1) on both the sides with respect to $x$.
Therefore we get
$\int {\dfrac{1}{y} \times y'dx = \int {\dfrac{x}{{{x^2} + 1}}dx} }$
Observing the above equation carefully, we can apply one of the integration formula which is
$\int {\dfrac{{x'}}{x}dx} = \left( {\ln x} \right)$
We can use this formula on Left hand side as well on the right hand side. But to apply this on the right hand side we will have to make minute changes that multiply and divide by 2 because the derivative of ${x^2} + 2$ is $2x$.
Making this changes we get
$\ln y = \dfrac{1}{2}\left( {\ln \left( {{x^2} + 1} \right)} \right) + C$ where $C$ is the constant of integration.
We can further write the above equation as
$\ln y = \dfrac{1}{2}\left( {\ln \left( {{x^2} + 1} \right)} \right) + \ln C$ since $\ln C$ is a constant
Further using the properties of log, we get
$\ln y = \dfrac{1}{2}\ln C\left( {{x^2} + 1} \right)$
Solving further we get
$2\ln y = \ln C\left( {{x^2} + 1} \right)$
$\ln {y^2} = \ln C\left( {{x^2} + 1} \right)$ (By using the property of log)
Taking antilog on both sides we get
${y^2} = C\left( {{x^2} + 1} \right)$
Now taking square root on both sides we get
$y = \sqrt {C\left( {{x^2} + 1} \right)}$

Hence the General solution to the given equation is $y = \sqrt {C\left( {{x^2} + 1} \right)}$

Note: A equation which is separable generally looks like $\dfrac{{dy}}{{dx}} = \dfrac{{g(x)}}{{f(y)}}$. But not always is our question given to us in this general form. If we try to conclude the equation by separating the variables the equation becomes easier to solve and these kinds of questions are generally known as variable separable differential equations.