Question-31485

Question Number 31485 by mondodotto@gmail.com last updated on 09/Mar/18

Answered by ajfour last updated on 09/Mar/18

(i) l e t y - 2 = t (x - 3)

\Rightarrow \frac{d y}{d x} = (x - 3) \frac{d t}{d x} + t

T h e n d i f f . e q . b e c o m e s

(x - 3) \frac{d t}{d x} + t = \frac{t (x - 3)}{(t + 1) (x - 3)}

\Rightarrow (x - 3) \frac{d t}{d x} + t = \frac{t}{t + 1}

o r (x - 3) \frac{d t}{d x} = \frac{t}{t + 1} - t

\int \frac{(t + 1) d t}{t^{2}} = - \int \frac{d x}{x - 3}

\Rightarrow \ln ∣ t ∣ - \frac{1}{t} = - \ln ∣ x - 3 ∣ + c

\Rightarrow \ln ∣ \frac{y - 2}{x - 3} ∣ + \ln ∣ x - 3 ∣= \frac{x - 3}{y - 2} + c

\Rightarrow \ln ∣ y - 2 ∣= \frac{x - 3}{y - 2} + c .

Answered by ajfour last updated on 09/Mar/18

(i i) l e t 2 x + y + 1 = t

\Rightarrow \frac{d y}{d x} = \frac{d t}{d x} - 2

D i f f e r e n t i a l e q . b e c o m e s

\frac{d t}{d x} - 2 = \frac{t - 3}{t}

\Rightarrow \frac{d t}{d x} = \frac{3 t - 3}{t}

o r \int \frac{t}{t - 1} d t = 3 \int d x

\Rightarrow \int d t + \int \frac{d t}{t - 1} = 3 x + c

o r t + \ln ∣ t - 1 ∣= 3 x + c

\Rightarrow 2 x + y + 1 + \ln ∣ 2 x + y ∣= 3 x + c .

Commented by mondodotto@gmail.com last updated on 09/Mar/18

thanx a lot .