Mathematics - Differential Equation Question with Solution | TestHub

$\left(x^2+xy+4x+2y+4\right)\frac{dy}{dx}-y^2=0$
⇒$\left({\left(x+2\right)}^2+y\left(x+2\right)\right)\frac{dy}{dx}=y^2$
Let$x+2=X, y=Y$
⇒$\left(X\right)\left(X+Y\right)\frac{dY}{dX}=Y^2$
⇒$-X^{2}dY=XYdY-Y^{2}dX$
⇒$-X^{2}dY=Y\left(XdY-YdX\right)$
⇒$-\frac{dY}{Y}=\frac{XdY-YdX}{X^2}$
Integrate both sides
⇒$-\mathcal{l}n\left|Y\right|=\left(\frac{Y}{X}\right)+C$
Putting back X, Y in x, y terms
⇒$-\mathcal{l}n\left|y\right|=\frac{y}{x+2}+C$
$\because$it is passing through (1, 3) ,$-\mathcal{ }\mathcal{l}n3=1+C$
$C= -1- \mathcal{l}n3$
$\therefore$curve equation is$\frac{y}{x+2}+\mathcal{ }\mathcal{l}n\left|y\right|-1- \mathcal{l}n3=0, x>0$.....(i)
Put$y=x+2$in equation (i)
then$\frac{x+2}{x+2}+\mathcal{ }\mathcal{l}n\left|x+2\right|-1- \mathcal{l}n 3=0$
$x=1, -5$(reject as given$x>0$)
$\therefore$curve intersect$y=x+2$at point (1, 3),(Only one point); put$y={\left(x+2\right)}^2$, we will get$x+2+2\mathcal{l}n\left(x+2\right)=1+ \mathcal{l}n3$
Clearly left hand side is an increasing function. Hence, it is always greater than$2+2\mathcal{l}n2$(By Putting X = 0)
Therefore no solution$y={\left(x+2\right)}^2$put$y={\left(x+3\right)}^2$in equation (i)
$\frac{{\left(x+3\right)}^2}{x+2}+\mathcal{l}n{\left(x+3\right)}^2-1-\mathcal{l}n3=0$
$\frac{{\left(x+3\right)}^2}{x+2}+\mathcal{l}n\frac{{\left(x+3\right)}^2}{3}-1=0$
$\because$$x>0 \Rightarrow x+3>x+2$and$x+3>3$
So$\frac{{\left(x+3\right)}^2}{x+2}+\mathcal{l}n\frac{{\left(x+3\right)}^2}{3}>1$
$\therefore$$\frac{{\left(x+3\right)}^2}{x+2}+\mathcal{l}n\frac{{\left(x+3\right)}^2}{3}-1=0$has no solution
$\Rightarrow$curve$y ={\left(x+3\right)}^2$does not intersect