Question

Find the solution of $(2x+3y-5)\frac{dy}{dx}+3x+2y-5=0$

• A. $3(x^2+y^2)-4xy+10(x+y)=k.$
• B. $3(x^2+y^2)+4xy+10(x+y)=k.$
• C. $3(x^2+y^2)-4xy-10(x+y)=k.$
• D. $3(x^2+y^2)+4xy-10(x+y)=k.$

Answer 1

The correct option is D $3(x^2+y^2)+4xy-10(x+y)=k.$

Given, $\frac{dy}{dx}=-\frac{3x+2y-5}{2x+3y-5}.$
Proceeding as usual, we get $\frac{3v+2}{3v^2+4v+3}dv=\frac{d\alpha }{\alpha }$, where $v=\frac{\beta }{\alpha }$
or $\frac{6v+4}{3v^2+4v+3}dv=-2\frac{d\alpha }{\alpha }.$
Integrating, $\log (3v^2+4v+3)=-2\log \alpha +c,$
$\log \frac{3{\beta }^2+4\alpha \beta +3{\alpha }^2}{{\alpha }^2}+\log {\alpha }^2=\log k.$
$3{\beta }^2+4\alpha \beta +3{\alpha }^2=k$ where $\beta =y-k=y-1$ and $\alpha =x-h=x-1.$
$3(x^2+y^2)+4xy-10(x+y)=k.$