A solution curve of differential equation ( x ^ 2 + xy + 4x + 2y +...

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