Question

Solve $x^2-3x+12=5$

Answer 1

$\frac{dy}{dx}=1+x+y+xy$

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

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

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

Now, integrating both sides,

$\int \frac{dy}{dx}=\int (x+1)dx$

$\Rightarrow$ $ln(y+1)=\frac{x^2}{2}+x+c$

Now, taking exponential both sides,

$\Rightarrow$ $y+1=e^{\frac{x^2}{2}+x+c}$

$\therefore$ $y=e^{\frac{x^2}{2}+x+c}-1$$x^2-3x+12=5$

$\Rightarrow x^2-3x+12=0$

$\Rightarrow x=\frac{3\pm \sqrt{{\left(-3\right)}^2-4\times 1\times 7}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{9-28}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{-19}}{2}$

$\Rightarrow x=\frac{3\pm \sqrt{19}i}{2}$