Question

For the given differential equation find the particular solution satisfying the given conditions.

(x+y)dy+(x-y)dx=0, y=1 when x=1.

Answer 1

Given, (x+y)dy+(x-y)dx=0 $\Rightarrow \frac{dy}{dx}=\frac{y-x}{x+y}$ ...(i)
Thus, the given differential equations is homogeneous.
So, put $\frac{y}{x}=vi.e.,$ $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Then, Eq. (i) becomes $v+x\frac{dv}{dx}=\frac{vx-x}{x+vx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{x(v-1)}{x(1+v)}\Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{1+v}$
$\Rightarrow x\frac{dv}{dx}=\frac{v-1}{1+v}-v\Rightarrow x\frac{dv}{dx}=\frac{v-1-v-v^2}{1+v}\Rightarrow -x\frac{dv}{dx}=\frac{1+v^2}{1+v}\Rightarrow \frac{(v+1)}{1+v^2}=-\frac{1}{x}dx$
On integrating both sides, we get
$\int \frac{(v+1)}{1+v^2}dv=-\int \frac{dx}{x}\Rightarrow \int \frac{v}{1+v^2}dv+\int \frac{1}{1+v^2}dv=-\int \frac{dx}{x}$
In first integarl, put
$v^2+1=t\Rightarrow 2v=\frac{dt}{dv}\Rightarrow dv=\frac{dt}{2v}\therefore \int \frac{v}{t}\times \frac{dt}{2v}+\int \frac{1}{1+v^2}dv=-\int \frac{dx}{x}\Rightarrow \frac{1}{2}\int \frac{1}{t}dt+\int \frac{1}{1+v^2}dv=-\int \frac{dx}{x}\Rightarrow \frac{1}{2}log|t|+ta{n}^{-1}v=-log|x|+C$ $[\because \int \frac{1}{1+x^2}dx=ta{n}^{-1}x]$
$\Rightarrow \frac{1}{2}log|v^2+1|+ta{n}^{-1}v=-log|x|+C$ $[\because t=v^2+1]$
$\Rightarrow \frac{1}{2}log\left(\frac{y^2+x^2}{x^2}\right)+log|x|+ta{n}^{-1}\left(\frac{y}{x}\right)=C$ $(Putv=\frac{y}{x})$
$\Rightarrow log\left(\frac{y^2+x^2}{x^2}\right)+2log|x|+2ta{n}^{-1}\left(\frac{y}{x}\right)=2C\Rightarrow log\left(\frac{y^2+x^2}{x^2}\right)+log{x}^2+2ta{n}^{-1}\left(\frac{y}{x}\right)=2C\Rightarrow log\frac{(y^2+x^2)x^2}{x^2}+2ta{n}^{-1}\left(\frac{y}{x}\right)=2C\Rightarrow log(x^2+y^2)+2ta{n}^{-1}\left(\frac{y}{x}\right)=A$ $(\because ta{n}^{-1}(1)=\frac{pi}{4})$
On putting the value of A in Eq. (ii), we get
$log(x^2+y^2)+2ta{n}^{-1}\left(\frac{y}{x}\right)=\frac{pi}{2}+log2$
This is the required solution of the given differential equation.