Question

solve (x+y+1)dy/dx=1​

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: \to \:(x + y + 1)\dfrac{dy}{dx} = 1$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{1}{x + y + 1} - - - (1)$

$\tt \: \to \:Put \: x + y + 1 = t$

$\tt \: \implies \: 1 + \dfrac{dy}{dx} + 0 = \dfrac{dt}{dx}$

$\tt \: \implies \: \dfrac{dy}{dx} = \dfrac{dt}{dx} - 1$

☆ So, (1) can be rewritten as

$\tt \: \implies \: \dfrac{dt}{dx} - 1 = \dfrac{1}{t}$

$\tt \: \implies \: \dfrac{dt}{dx} \: = \dfrac{t + 1}{t}$

$\tt \: \implies \: \dfrac{t}{t + 1} dt \: = dx$

☆ On integrating, both sides we get

$\tt \: \to \: \int \: \dfrac{t}{t + 1} dt = \int \: dx$

$\tt \: \to \: \int \: \dfrac{t + 1 - 1}{t + 1}dt = x + c$

$\tt \: \to \: \: \int \: (1 - \dfrac{1}{t + 1} ) = x + c$

$\tt \: \to \: t - log(t + 1) = x + c$

$\tt \: \to \:x + y + 1 - log(x + y + 1+ 1) = x + c$

$\tt \: \to \:y + 1 - log(x + y + 2) = c$