Question

(3x+y-5)dy = (2x+2y-2)dx solve this nonhomogeneous differential equation ... don't spam then i will report your answer and account ....... ✌️✌️✨✨​

Answer 1

$\left(3x+y-5\right)dy=\left(2x+2y-2\right)dx:\quad -\frac{1}{3}\ln \left(\frac{2x+2y-2}{3x+y-5}-1\right)+\frac{1}{12}\ln \left(\frac{2x+2y-2}{3x+y-5}+2\right)+\frac{1}{4}\ln \left(\frac{2x+2y-2}{3x+y-5}-2\right)=\frac{1}{4}\ln \left(4x-8\right)+c_1$

$= \left(3x+y-5\right)\frac{dy}{dx}=2x+2y-2$

$= \left(3x+y-5\right)y'\:=2x+2y-2$

$= \left(\frac{2x-2-3xv+5v}{v-2}\right)'\:=\frac{2x+2\cdot \frac{2x-2-3xv+5v}{v-2}-2}{3x+\frac{2x-2-3xv+5v}{v-2}-5}$

$= \left(\frac{2x-2-3xv+5v}{v-2}\right)'\:=v$

$= \left(\frac{2x-2-3xv+5v}{v-2}\right)'\:=\frac{4xv'\:-8v'\:+8v-3v^2-4}{\left(v-2\right)^2}$.

$= \frac{4xv'\:-8v'\:+8v-3v^2-4}{\left(v-2\right)^2}=v$

$= \frac{4xv'\:-8v'\:+8v-3v^2-4}{\left(v-2\right)^2}=v:\quad -\frac{1}{3}\ln \left(v-1\right)+\frac{1}{12}\ln \left(v+2\right)+\frac{1}{4}\ln \left(v-2\right)=\frac{1}{4}\ln \left(4x-8\right)+c_1$

$= -\frac{1}{3}\ln \left(\frac{2x+2y-2}{3x+y-5}-1\right)+\frac{1}{12}\ln \left(\frac{2x+2y-2}{3x+y-5}+2\right)+\frac{1}{4}\ln \left(\frac{2x+2y-2}{3x+y-5}-2\right)=\frac{1}{4}\ln \left(4x-8\right)+c_1$

Answer 2

Answer:

REFER TO THE ATTACHMENT