Question

Solution of $\frac{dy}{dx}=4+4x-3y-3xy$ is:

• A. $2\log (4-3y)+3x^2+6x=c$
• B. $\log (3-4y)+3y^2+6y=c$
• C. $(4-3y)(1+x)=c$
• D. $\log (4-3y)+x^2+3x=c$

Answer 1

The correct option is A $2\log (4-3y)+3x^2+6x=c$

$\frac{dy}{dx}=4(1+x)-3y(1+x)$

$\Rightarrow \int \frac{dy}{4-3y}=\int (1+x)dx+c$

$\Rightarrow -\frac{1}{3}log(\frac{4}{3}-y)=x+\frac{x^2}{2}+c$

$\Rightarrow c=\frac{2x+x^2}{2}+\frac{1}{3}log\left(\frac{4-3y}{3}\right)$

$\Rightarrow c=6x+3x^2+2log(4-3y)$