Question

The general solution of the D.E. ${{x}^{2}}\dfrac{dy}{dx}={{x}^{2}}+xy+{{y}^{2}}$ is:
$\left( A \right)\text{ }{{\tan }^{-1}}\dfrac{y}{x}=\log x+c$
$\left( B \right)\text{ }{{\tan }^{-1}}\dfrac{x}{y}=\log x+c$
$\left( C \right)\text{ }{{\tan }^{-1}}\dfrac{y}{x}=\log y+c$
$\left( D \right)\text{ none of these}$

Answer 1

Hint: In this question we have been given a differential equation which is in the homogenous form. We will solve this equation by substituting the value of $y$ as $vx$ where $v$ is a multiple that converts the $x$ terms into the $y$ term. We will then differentiate the equation and substitute it in the given equation. We will then take out terms common and simplify and separate the variables and integrate them to get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow {{x}^{2}}\dfrac{dy}{dx}={{x}^{2}}+xy+{{y}^{2}}\to \left( 1 \right)$
We can see that the given differential equation is a homogeneous differential equation and the degree of all the terms is $2$ therefore, we substitute $y=vx$, we have:
$\Rightarrow y=vx$ therefore, we have $v=\dfrac{y}{x}\to \left( 2 \right)$
On differentiating with respect to $x$, we get:
$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}\to \left( 3 \right)$
On substituting equations $\left( 2 \right)$ and $\left( 3 \right)$ in $\left( 1 \right)$, we get:
$\Rightarrow {{x}^{2}}\left( v+\dfrac{dv}{dx} \right)={{x}^{2}}+x\left( vx \right)+{{\left( vx \right)}^{2}}$
On simplifying, we get:
$\Rightarrow {{x}^{2}}\left( v+x\dfrac{dv}{dx} \right)={{x}^{2}}+v{{x}^{2}}+{{v}^{2}}{{x}^{2}}$
On taking the term ${{x}^{2}}$ common in the right-hand side, we get:
$\Rightarrow {{x}^{2}}\left( v+x\dfrac{dv}{dx} \right)={{x}^{2}}\left[ 1+v+{{v}^{2}} \right]$
On cancelling the term ${{x}^{2}}$ from both the sides, we get:
$\Rightarrow v+x\dfrac{dv}{dx}=1+v+{{v}^{2}}$
On cancelling the term $v$, we get:
$\Rightarrow x\dfrac{dv}{dx}=1+{{v}^{2}}$
On transferring the term $1+{{v}^{2}}$ from the right-hand side to the left-hand side, we get:
$\Rightarrow \dfrac{x\cdot dv}{1+{{v}^{2}}\cdot dx}=1$
On transferring the terms $x$ $dx$ from the left-hand side to the right-hand side, we get:
$\Rightarrow \dfrac{dv}{1+{{v}^{2}}}=\dfrac{dx}{x}$
Now since all the variables are separated, we will integrate the expression therefore, we have:
$\Rightarrow \int{\dfrac{dv}{1+{{v}^{2}}}}=\int{\dfrac{dx}{x}}$
Now we know that $\int{\dfrac{dv}{1+{{v}^{2}}}}={{\tan }^{-1}}\left( v \right)+c$ and $\int{\dfrac{dx}{x}}=\log x+c$, we get:
$\Rightarrow {{\tan }^{-1}}\left( v \right)=\log x+c$
Now from equation $\left( 2 \right)$, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{y}{x} \right)=\log x+c$

So, the correct answer is “Option A”.

Note: It is to be remembered that there are various forms of differential equations and in this question, we have a homogenous equation. The basic method of differential equations should be remembered by separating the variables on the opposite sides of the equation and then integrating both the sides.