Question

Solve:$(x-1)\frac{dy}{dx}{x}^2+3x+2$

Answer 1

$(x-1)\frac{dy}{dx}{x}^2+3x+2=0\Rightarrow dy=\frac{-(3x+2)}{(x-1)x^2}dx\Rightarrow \int dy=-\int \frac{(3x+2)}{(x-1)x^2}dx$

By partial fraction;

$\frac{3x+2}{(x-1)x^2}=\frac{A}{x-1}+\frac{Bx+C}{x^2}\Rightarrow 3x+2=(A+B)x^2-(B-C)x-C$

comparing both side we get;

$A+B=0B-C=-3c=-2$

from above equations we get

$A=5;B=-5$

$\therefore \int dy=-\int \frac{3x+2}{(x-1)x^2}dx\Rightarrow y=-\int [\frac{5}{x-1}+\left(\frac{-5x-2}{x^2}\right)]dx\Rightarrow y=-\int [\frac{5}{x-1}-\frac{5}{x}-\frac{2}{x^2}]dx\Rightarrow y=-\left[5\log \right(x-1)-5\log x+\frac{2}{x}]+C\Rightarrow y=\log x^5-\log {(x-1)}^5-\frac{2}{x}+C\Rightarrow y=\log \frac{x^5}{{(x-1)}^5}-\frac{2}{x}+C$