Question

For the differential equations, find the particular solution satisfying the given condition:\([xsin^2(\frac{x}{y})-y]dx+xdy=0;y=\frac{π}{4},\)when \(x=1\)

Answer 1

\([xsin^2(\frac{x}{y})-y]dx+xdy=0\)
\(⇒\frac{dy}{dx}=\frac{-[xsin^2(\frac{y}{x})-y]}{x}...(1)\)
Let \(F(x,y)=\frac{-[xsin^2(\frac{y}{x})-y]}{x}\)
\(∴F(λx,λy)=[λx.sin^2(\frac{λx}{λy})-λy]=\frac{-[xsin^2(\frac{y}{x})-y]}{x}=λ°.F(x,y)\)
Therefore,the given differential equation is a homogenous equation.
To solve this differential equation,we make the substitution as:
\(y=vx\)
\(⇒\frac{d}{dx}(y)=\frac{d}{dx}(vx)\)
\(⇒\frac{dy}{dx}=v+x=\frac{dv}{dx}\)
Substituting the values of y and \(\frac{dy}{dx}\) in equation(1),we get:
\(v+x\frac{dv}{dx}=\frac{-[xsin^2v-vx]}{x}\)
\(⇒v+x\frac{dv}{dx}=-[sin^2v-v]=v-sin^2v\)
\(⇒x\frac{dv}{dx}=-sin^2v\)
\(⇒\frac{dv}{sin^2v}=\frac{-dx}{dx}\)
\(⇒cosec^2\,v\,dv=\frac{-dx}{x}\)
\(-cotv=-log|x|-C\)
\(⇒cotv=log|x|+C\)
\(⇒cot(\frac{y}{x})=log|x|+log|C|\)
\(⇒cot(\frac{y}{x})=log|Cx|...(2)\)
Now,\(y=\frac{π}{4}\) at x=1
\(⇒cot(\frac{π}{4})=log|C|\)
\(⇒1=logC\)
\(⇒C=e^1=e\)
Substituting \(C=e\) in equation(2),we get:
\(cot(\frac{y}{x})=log|ex|\)
This is the required solution of the given differential equation.