Question

Find the solution of the differential equation $ {{x}^{2}}dy+y\left( x+y \right)dx=0 $ , if $ x=1 $ and $ y=1 $ .

Answer 1

Hint: First, we will find $ \dfrac{dy}{dx} $ from the equation $ {{x}^{2}}dy+y\left( x+y \right)dx=0 $ . We will get as $ \dfrac{dy}{dx}=-\dfrac{y\left( x+y \right)}{{{x}^{2}}} $ . Then, we will assume $ \dfrac{y}{x}=v $ and on differentiating e will find value of $ \dfrac{dy}{dx} $ in terms of variable v. Then we will be doing integration in order to get value of constant term and finally on further simplification we will get the required solution of differential equation. Formula we will use are $ \dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u $ , $ \int{\dfrac{1}{x}dx=\log x} $ , $ m\log n=\log {{n}^{m}} $ .

Complete step-by-step answer:
Here, we will first find $ \dfrac{dy}{dx} $ from the equation $ {{x}^{2}}dy+y\left( x+y \right)dx=0 $ . So, we can write it as
 $ {{x}^{2}}dy=-y\left( x+y \right)dx $
Further, we will take dx to left hand side in denominator and $ {{x}^{2}} $ to right hand side in denominator, we get as
 $ \dfrac{dy}{dx}=-\dfrac{y\left( x+y \right)}{{{x}^{2}}} $
Now, further multiplying the brackets we can write it as
 $ \dfrac{dy}{dx}=-\left( \dfrac{yx+{{y}^{2}}}{{{x}^{2}}} \right) $
 $ \dfrac{dy}{dx}=-\left( \dfrac{yx}{{{x}^{2}}}+\dfrac{{{y}^{2}}}{{{x}^{2}}} \right) $
Now, we will cancel the similar terms in numerator and denominator we can write it as
 $ \dfrac{dy}{dx}=-\left( \dfrac{y}{x}+{{\left( \dfrac{y}{x} \right)}^{2}} \right) $ …………………………(1)
Now, we will assume that let $ \dfrac{y}{x}=v $ . So, we can write it as $ y=vx $ . On differentiating it with respect to x, we get as
 $ \dfrac{dy}{dx}=v\dfrac{d}{dx}x+x\dfrac{dv}{dx} $ (using product rule $ \dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u $ )
Further, on solving we get as
 $ \dfrac{dy}{dx}=v+x\dfrac{dv}{dx} $
Now, we will substitute the above value in equation (1) and also $ \dfrac{y}{x}=v $ . We will get as
 $ v+x\dfrac{dv}{dx}=-\left( v+{{v}^{2}} \right) $
On solving we can write it as
 $ x\dfrac{dv}{dx}=-v-v-{{v}^{2}}=-2v-{{v}^{2}} $
 $ x\dfrac{dv}{dx}=-v\left( v+2 \right) $
Now, taking v terms on the left hand side and x terms on the right hand side. We will get as
 $ \dfrac{dv}{v\left( v+2 \right)}=-\dfrac{dx}{x} $
We will take integration on both sides of the equation.
 $ \int{\dfrac{dv}{v\left( v+2 \right)}}=-\int{\dfrac{dx}{x}} $ …………………………….(2)
There is no formula for $ \int{\dfrac{dv}{v\left( v+2 \right)}} $ so, we will find numerator such that $ \dfrac{1}{v\left( v+2 \right)}=\dfrac{A}{v}+\dfrac{B}{\left( v+2 \right)} $
On taking LCM, we get as
 $ \dfrac{1}{v\left( v+2 \right)}=\dfrac{A\left( v+2 \right)+Bv}{v\left( v+2 \right)} $
On cancelling the denominator terms, and assuming v to be 0, we get value of A to be
 $ 1=A\left( v+2 \right)+Bv $
 $ 1=2A\Rightarrow A=\dfrac{1}{2} $ …………………..(3)
Similarly, if we put V as $ -2 $ , we get B as
 $ 1=A\left( -2+2 \right)+-2B $
 $ 1=-2B\Rightarrow B=-\dfrac{1}{2} $ ………………….(4)
Thus, we can write $ \dfrac{1}{v\left( v+2 \right)}=\dfrac{1}{2v}-\dfrac{1}{2\left( v+2 \right)} $ . So, we will substitute the values in equation (2). So, we will get as
 $ \int{\dfrac{1}{2v}dv-\int{\dfrac{1}{2\left( v+2 \right)}}}dv=-\int{\dfrac{1}{x}dx} $
We know that $ \int{\dfrac{1}{x}dx=\log x} $ . Using this we will get as
 $ \dfrac{1}{2}\log v-\dfrac{1}{2}\log \left( v+2 \right)=-\log x+\log c $ ( \[\log c\] is constant term)
Now, we will use the formula $ m\log n=\log {{n}^{m}} $ , \[\log m-\log n=\log \left( \dfrac{m}{n} \right)\] . So, we can write equation as
\[\dfrac{1}{2}\left[ \log v-\log \left( v+2 \right) \right]=\log \dfrac{c}{x}\]
\[\log \left( \dfrac{v}{v+2} \right)=2\log \dfrac{c}{x}\]
\[\log \left( \dfrac{v}{v+2} \right)=\log {{\left( \dfrac{c}{x} \right)}^{2}}\]
Cancelling log on both sides and putting the value of v which we assumed above. We will get as
\[\left( \dfrac{v}{v+2} \right)={{\left( \dfrac{c}{x} \right)}^{2}}\]
\[{{x}^{2}}\left( \dfrac{\dfrac{y}{x}}{\dfrac{y}{x}+2} \right)={{c}^{2}}\]
On further simplifying, we get as
\[{{x}^{2}}\left( \dfrac{\dfrac{y}{x}}{\dfrac{y+2x}{x}} \right)={{c}^{2}}\]
On cancelling the denominator and further solving, we get as
\[\left( \dfrac{{{x}^{2}}y}{y+2x} \right)={{c}^{2}}\] ……………….(5)
Here, we will assume a constant term to be let say k. And we are given that $ x=1 $ and $ y=1 $ . By putting this, we will find the value of k. We will get as
\[\left( \dfrac{1}{1+2} \right)=k=\dfrac{1}{3}\]
On putting value in equation (5) on solving we will get answer as
\[\left( \dfrac{{{x}^{2}}y}{y+2x} \right)=\dfrac{1}{3}\]
\[3{{x}^{2}}y=y+2x\] …………………….(6)
Thus, the solution of differential equation $ {{x}^{2}}dy+y\left( x+y \right)dx=0 $ , if $ x=1 $ , $ y=1 $ is \[3{{x}^{2}}y=y+2x\] .

Note: Students should know the formula related to integration and be careful while solving this type of equation. There are chances of calculation errors. Also, as $ \dfrac{dy}{dx} $ is given in equation form do not directly put $ x=1 $ , $ y=1 $ in the equation $ {{x}^{2}}dy+y\left( x+y \right)dx=0 $ and calculate value of $ \dfrac{dy}{dx} $ . This is not the correct way to get an answer. We have to find the value of the constant term using the data given. So, be careful and avoid making mistakes.