Question

For the differential equations, find the particular solution satisfying the given condition:\((x+y)dy+(x-y)dy=0;y=1\) when \(x=1\)

Answer 1

\((x+y)dy+(x-y)dy=0\)
\(⇒\frac{dy}{dx}=\frac{-(x-y)}{x+y}...(1)\)
Let \(F(x,y)=\frac{-(x-y)}{x+y}.\)
\(∴F(λx,λy)=\frac{-(λx-λy)}{λx-λy}=\frac{-(x-y)}{x+y}=λ0.F(x,y)\)
Therefore,the given differential equation is a homogenous equation.
To solve it,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:
\(v+x\frac{dv}{dx}=\frac{-(x-vx)}{x+vx}\)
\(⇒v+x\frac{dv}{dx}=\frac{v-1}{v+1}\)
\(⇒x\frac{dv}{dx}=\frac{v-1}{v+1}-v=\frac{v-1-v(v+1)}{v+1}\)
\(⇒x\frac{dv}{dx}=\frac{v-1-v^2-v}{v+1}=\frac{-(1+v^2)}{v+1}\)
\(⇒\frac{(v+1)}{1+v^2}dv=\frac{-dx}{x}\)
\(⇒[\frac{v}{1+v^2}+\frac{1}{1+v^2}]dv=\frac{-dx}{x}\)
Integrating both sides,we get:
\(\frac{1}{2}log(1+v^2)+tan^{-1}v=-logx+k\)
\(⇒log(1+v^2)+tan^{-1}v=-logx+k\)
\(⇒log[(1+v^2).x^2]+2tan^{-1}v=2k\)
\(⇒log[(1+\frac{y^2}{x^2}).x^2]+2tan^{-1}\frac{y}{x}=2k\)
\(⇒log(x^2+y^2)+2tan^{-1}\frac{y}{x}=2k...(2)\)
Now,\(y=1\) at \(x=1.\)
\(⇒log2+tan^{-1}=2k\)
\(⇒log2+2\times\frac{π}{4}=2k\)
\(⇒\frac{π}{4}+log2=2k\)
Substituting the value of 2k in equation(2),we get:
\(log(x^2+y^2)+2tan{-1}(\frac{y}{x})=\frac{π}{4}+log2\)
This is the required solution of the given differential equation.