Solve-the-differential-equation-y-2x-3y-4-4x-3y-2-

Question Number 14444 by tawa tawa last updated on 31/May/17

Solve the differential equation

y^{'} = \frac{2 x + 3 y - 4}{4 x + 3 y + 2}

Answered by mrW1 last updated on 31/May/17

l e t x = u + p a n d y = v + q

2 x + 3 y - 4 = 2 u + 3 v + 2 p + 3 q - 4

4 x + 3 y + 2 = 4 u + 3 v + 4 p + 3 q + 2

l e t

2 p + 3 q - 4 = 0 \dots (i)

4 p + 3 q + 2 = 0 \dots (i i)

(i i) - (i) :

2 p + 6 = 0

\Rightarrow p = - 3

2 (i) - (i i) :

3 q - 10 = 0

\Rightarrow q = \frac{10}{3}

\Rightarrow x = u - 3

\Rightarrow y = v + \frac{10}{3}

d x = d u

d y = d v

\Rightarrow \frac{d v}{d u} = \frac{2 u + 3 v}{4 u + 3 v}

l e t v = t u

\frac{d v}{d u} = t + u \frac{d t}{d u}

\Rightarrow t + u \frac{d t}{d u} = \frac{2 + 3 t}{4 + 3 t}

u \frac{d t}{d u} = \frac{2 + 3 t}{4 + 3 t} - t = - \frac{3 t^{2} - t - 2}{4 + 3 t} = - \frac{(3 t + 2) (t - 1)}{3 t + 4}

\frac{3 t + 4}{(3 t + 2) (t - 1)} d t = - \frac{d u}{u}

\frac{3 t + 2 + 2}{(3 t + 2) (t - 1)} d t = - \frac{d u}{u}

[\frac{3}{5} (\frac{1}{t - 1}) - \frac{6}{5} (\frac{1}{3 t + 2})] d t = - \frac{d u}{u}

\int [\frac{3}{5} (\frac{1}{t - 1}) - \frac{6}{5} (\frac{1}{3 t + 2})] d t = - \int \frac{d u}{u}

\frac{3}{5} \int \frac{1}{t - 1} d t - \frac{6}{5} \int \frac{1}{3 t + 2} d t = - \int \frac{d u}{u}

\frac{3}{5} \ln (t - 1) - \frac{18}{5} \ln (3 t + 2) = - \ln u + C_{1}

\frac{3}{5} \ln (\frac{v}{u} - 1) - \frac{18}{5} \ln (\frac{3 v}{u} + 2) = - \ln u + C_{1}

\frac{3}{5} \ln (\frac{y - \frac{10}{3}}{x + 3} - 1) - \frac{18}{5} \ln (\frac{3 y - 10}{x + 3} + 2) = - \ln (x + 3) + C_{1}

\frac{3}{5} \ln (\frac{y - x - 3}{x + 3}) - \frac{18}{5} \ln (\frac{3 y + 2 x - 4}{x + 3}) = - \ln (x + 3) + C_{1}

3 \ln (\frac{y - x - 3}{x + 3}) + 5 \ln (x + 3) - 18 \ln (\frac{3 y + 2 x - 4}{x + 3}) = C

3 \ln (y - x - 3) + 20 \ln (x + 3) - 18 \ln (3 y + 2 x - 4) = C

o r

\frac{{(y - x - 3)}^{3} {(x + 3)}^{20}}{{(3 y + 2 x - 4)}^{18}} = C

Commented by tawa tawa last updated on 31/May/17

God bless you sir . I really appreciate your effort .