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.