Question

Solve the following differential equation $\text{cosec}x\log y\dfrac{dy}{dx}+{{x}^{2}}{{y}^{2}}=0$.

Answer 1

In this question, in order to solve the given differential equation $\text{cosec}x\log y\dfrac{dy}{dx}+{{x}^{2}}{{y}^{2}}=0$ we have to simplify it using variable separable to get $\dfrac{\log y}{{{y}^{2}}}dy=\dfrac{-{{x}^{2}}}{\text{cosec}x}dx$ . We will then integrate both sides to get the desired solution of the given differential equation.

Complete step by step answer:
We are given with a differential equation of the form $\text{cosec}x\log y\dfrac{dy}{dx}+{{x}^{2}}{{y}^{2}}=0$.
On taking the value ${{x}^{2}}{{y}^{2}}$ on the right hand side of the above differential equation, we get
$\text{cosec}x\log y\dfrac{dy}{dx}=-{{x}^{2}}{{y}^{2}}$.
Now we will be separating the variables by taking terms in variable $x$ on the right hand side of the above equation and terms in variable $y$ on the left hand side of the above equation.
Then we get
$\dfrac{\log y}{{{y}^{2}}}dy=\dfrac{-{{x}^{2}}}{\text{cosec}x}dx$

Now on integrating both sides of the above differential equation, we will have
$\int{\dfrac{\log y}{{{y}^{2}}}dy}=\int{\dfrac{-{{x}^{2}}}{\text{cosec}x}dx}$
Since $\dfrac{1}{\text{cosec}x}=\sin x$, then the above equation becomes
$\int{\log y\dfrac{1}{{{y}^{2}}}dy}=-\int{{{x}^{2}}\sin xdx}$
Since we know that by integration by parts we have $\int{\left( uv \right)dx=u\int{vdx-\int{\dfrac{d}{dx}\left( u \right)\int{vdx}}}}$
Now using formula of integration by parts on both side of the above integral, we get
$\log y\int{\dfrac{1}{{{y}^{2}}}dy-\int{\dfrac{d}{dy}\left( \log y \right)\int{\dfrac{1}{{{y}^{2}}}dy=}}}-\left[ {{x}^{2}}\int{\sin xdx-\int{\dfrac{d}{dx}\left( {{x}^{2}} \right)\int{\sin xdx}}} \right]$
Now on evaluating the integral, we will have
$\log y\left[ -\dfrac{1}{y} \right]-\int{\left( \dfrac{1}{y} \right)\left[ -\dfrac{1}{y} \right]dy=}{{x}^{2}}\cos x-\int{\left( 2x \right)\cos xdx}+c$
On simplifying the above equation we get
$-\dfrac{\log y}{y}+\int{\left( \dfrac{1}{{{y}^{2}}} \right)dy=}{{x}^{2}}\cos x-2\int{x\cos xdx}+c$
Again on evaluation the above integral by using integration by parts in the right hand side of the above equation we get,
$-\dfrac{\log y}{y}+\left[ -\dfrac{1}{y} \right]={{x}^{2}}\cos x-2\left[ x\int{\cos xdx}-\int{\dfrac{d}{dx}\left( x \right)\int{\cos xdx}} \right]+c$
Now we get
$-\dfrac{\log y}{y}+-\dfrac{1}{y}={{x}^{2}}\cos x-2\left[ x\sin x-\int{1.\sin xdx} \right]+c$
Again on simplifying, we will have
$-\dfrac{\log y}{y}-\dfrac{1}{y}={{x}^{2}}\cos x-2\left[ x\sin x+\cos x \right]+c$
$\Rightarrow \dfrac{\log y}{y}+\dfrac{1}{y}+{{x}^{2}}\cos x-2\left[ x\sin x+\cos x \right]+c=0$
There on solving the given differential equation $\text{cosec}x\log y\dfrac{dy}{dx}+{{x}^{2}}{{y}^{2}}=0$, we get $\dfrac{\log y}{y}+\dfrac{1}{y}+{{x}^{2}}\cos x-2\left[ x\sin x+\cos x \right]+c=0$.

Note:
In this problem, we are using the method of separation of variables of a differential equation in order to solve the problem. While using this method we have to take care of the fact that the differential equation allows one to rewrite an equation so that each of two variables occurs on a different side of the equation. Then integrating both sides with respect to different variables will lead us to the solution.