Question

Show that the differential equation $\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]dx+xdy=0$ is homogeneous. Find the particular solution of this differential equation, given that $y=\dfrac{\pi }{4}$ when $x=1$.

Answer 1

Hint: Homogeneous differential equation can be represented in the form $\dfrac{dy}{dx}=f\left( \dfrac{y}{x} \right)$ , so we convert the given equation in the above form to show that the equation given is homogeneous. Then, to find the particular solution of this differential equation we put $y=vx$ and calculate the value and integrate the obtained equation to get the desired solution.

Complete step by step answer:
We have been given the differential equation $\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]dx+xdy=0$
We have to show that the given equation is homogeneous and find the particular solution of this differential equation.
Now, first of all we need to check that the given equation is homogeneous or not.
We have $\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]dx+xdy=0$
We divide the whole equation by $dx$, we get
\[\begin{align}
  & \Rightarrow \dfrac{\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]dx}{dx}+\dfrac{xdy}{dx}=0 \\
 & \Rightarrow \left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]+\dfrac{xdy}{dx}=0 \\
 & \Rightarrow \dfrac{xdy}{dx}=-\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right] \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{\left[ x{{\sin }^{2}}\left( \dfrac{y}{x} \right)-y \right]}{x} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-x{{\sin }^{2}}\left( \dfrac{y}{x} \right)}{x}+\dfrac{y}{x} \\
 & \Rightarrow \dfrac{dy}{dx}=-{{\sin }^{2}}\left( \dfrac{y}{x} \right)+\dfrac{y}{x}............(i) \\
\end{align}\]
Now, we know that the general form of a homogeneous equation is $\dfrac{dy}{dx}=f\left( \dfrac{y}{x} \right)$, so when we compare the above equation with this general form we get that the above equation is homogeneous equation.
Now, we have to find the particular solution of this equation.
We put $y=vx$, where $v$ is a variable.
So, we have $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$
Now, substituting the values in equation (i) we have
\[\begin{align}
  & \Rightarrow v+x\dfrac{dv}{dx}=-{{\sin }^{2}}\left( \dfrac{vx}{x} \right)+\dfrac{vx}{x} \\
 & \Rightarrow v+x\dfrac{dv}{dx}=-{{\sin }^{2}}v+v \\
\end{align}\]
Simplifying further, we get
\[\begin{align}
  & \Rightarrow x\dfrac{dv}{dx}=-{{\sin }^{2}}v \\
 & \Rightarrow \dfrac{dv}{{{\sin }^{2}}v}=-\dfrac{dx}{x} \\
\end{align}\]
Now, we have separable variables, so we integrate the above equation.
\[\Rightarrow \int{\dfrac{dv}{{{\sin }^{2}}v}}=-\int{\dfrac{dx}{x}}\]
Or we can write \[\Rightarrow \int{cose{{c}^{2}}vdv}=-\int{\dfrac{dx}{x}}\]
Now, we know that $\int{\dfrac{1}{x}dx=\log x}$ and \[\int{cose{{c}^{2}}vdv=-\cot v}\]
Now, substituting the values, we get
\[\Rightarrow -\cot v=-\log x+c\]
Now, put $v=\dfrac{y}{x}$ , we get
\[\Rightarrow -\cot \dfrac{y}{x}=-\log x+c\]
Now, we have given $y=\dfrac{\pi }{4}$ when $x=1$.
Substituting the value, we get
\[\Rightarrow -\cot \dfrac{\dfrac{\pi }{4}}{1}=-\log 1+c\]
Now, we know that $\log 1=0$ and $\cot \dfrac{\pi }{4}=1$
So, we have
\[\begin{align}
  & \Rightarrow -1=0+c \\
 & \Rightarrow c=-1 \\
\end{align}\]

So, the particular solution of the given equation is
$\begin{align}
  & -\cot \dfrac{y}{x}=-\log x-1 \\
 & \cot \dfrac{y}{x}-\log x=1 \\
\end{align}$

Note: If the points are not given in the question then the general solution of the above equation is given by
\[\Rightarrow -\cot \dfrac{y}{x}=-\log x+c\]
Or we can write \[\Rightarrow \log x-\cot \dfrac{y}{x}=c\]