Question

Reduce each of the following differential equations to the variables separable form and hence solve . $i$) $1+\frac{dy}{dx}=cosec\left(x+y\right)$
$ii$) ${\left(x-y\right)}^2\frac{dy}{dx}=a^2$

Answer 1

$\left(i\right)$ $1+\frac{dy}{dx}=coses(x+y)$

let $x+y=t$

$1+\frac{dy}{dx}=\frac{dt}{dx}$

$\therefore \frac{dt}{dx}=cosect$

$\int dx=\int \sin tdt$

$x=-\cos t+c$

$x=-\cos (x+y)+c$

$\left(ii\right)$ $(x-y{)}^2\frac{dy}{dx}=a^2$

let $(x-y)=t$

$\therefore 1-\frac{dy}{dx}=\frac{dt}{dx}$

So we can write

$t^2(1-\frac{dt}{dx})=a^2$

$1-\frac{dt}{dx}=\frac{a^2}{t^2}$

$1-\frac{a^2}{t^2}=\frac{dt}{dx}$

$\int dx=\int \frac{1}{1-\frac{a^2}{t^2}}-dt$

$\int dx=\int \frac{t^2}{t^2-a^2}dt$

$x=\int \frac{(t^2-a^2)+a^2}{t^2-a^2}dt$

$x=\int dt+a^2\int \frac{1}{t^2-a^1}dt$

$x=t+a^2\frac{1}{2a}\log |\frac{t-a}{t+a}|+c$

$x=t+\frac{a}{2}\log |\frac{t-a}{t+a}|+c$

$x=(x-y)+\frac{a}{2}\log |\frac{x-y-a}{x-y+a}|+c$