Question

A solution curve of the differential equation $({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2} = 0,x > 0,$ passes through the point (1, 3). Then the solution curve
$
  {\text{(a) intersects }}y = x + 2{\text{ exactly at one point}} \\
  {\text{(b) intersects }}y = x + 2{\text{ exactly at two points}} \\
  {\text{(c) intersects }}y = {(x + 2)^2} \\
  {\text{(d) does NOT intersect }}y = {(x + 3)^2} \\
 $

Answer 1

Hint: For this problem, first to find out the solution curve with variable separable method for the given differential equation and put the values of $(x,y)$ which is given in the question, i.e., (1, 3). Then check each option by putting the value of $y$ in the solution curve.

Complete step-by-step answer:
Firstly, write the expression given in the question as,
$({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2} = 0,x > 0,$
This can be rearranged in the following way,
$
   \Rightarrow ({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2} = 0 \\
   \Rightarrow ({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} = {y^2} \\
   \Rightarrow {y^2}\dfrac{{dx}}{{dy}} = ({x^2} + xy + 4x + 2y + 4) \\
    \\
 $(Taking $\dfrac{{dy}}{{dx}}$ to other side of the equation, it will be $\dfrac{{dx}}{{dy}}$)
Now, factorising the above expression, we will get
$
   \Rightarrow {y^2}\dfrac{{dx}}{{dy}} = ({x^2} + 4x + 4) + y(x + 2) \\
   \Rightarrow {y^2}\dfrac{{dx}}{{dy}} = {(x + 2)^2} + y(x + 2) \\
 $
Now, divide the whole expression by ${y^2}$, we get
$ \Rightarrow \dfrac{{dx}}{{dy}} = {(\dfrac{{x + 2}}{y})^2} + (\dfrac{{x + 2}}{y}){\text{ }}.........{\text{(1)}}$
Now, put $z = (\dfrac{{x + 2}}{y})$ we get
$
  z = (\dfrac{{x + 2}}{y}) \\
  yz = x + 2 \\
 $
And, differentiating the above equation with respect to $y$, we get
$z + y\dfrac{{dz}}{{dy}} = \dfrac{{dx}}{{dy}}{\text{ }}$
And replacing the value of $\dfrac{{dx}}{{dy}}$ in eq. (1), we get
\[
  z + y\dfrac{{dz}}{{dy}} = {z^2} + z \\
  y\dfrac{{dz}}{{dy}} = {z^2} \\
  \dfrac{1}{{{z^2}}}dz = \dfrac{1}{y}dy,{\text{ now integrating,}} \\
  \int {\dfrac{1}{{{z^2}}}dz = } \int {\dfrac{1}{y}dy} \\
 \]
$ - \dfrac{1}{z} = \log y + c,$By putting the value of $z$in this equation, we will get
$ - \dfrac{y}{{x + 2}} = \log y + c$.Value of $c$at point (1, 3) is
$
  c = - 1 - {\text{log}}3 \\
   - \dfrac{y}{{x + 2}} = \log y - 1 - {\text{log}}3 \\
 $
Rearranging the above expression and taking out (balancing) the negative sign, we get
$
  \dfrac{y}{{x + 2}} + \log y = 1 + {\text{log}}3 \\
  \dfrac{y}{{x + 2}} + \log y = {\text{log}}3e{\text{ }}........{\text{ (2)}} \\
 $
Case 1: Checking option (a), by putting $y = x + 2$ in eq. (2), we get a unique value for $x$ which is
$
  \dfrac{{x + 2}}{{x + 2}} + \log (x + 2) = {\text{log}}3e \\
  1 + \log (x + 2) = {\text{log}}3e \\
  \log (x + 2) = {\text{log}}3e - 1 = {\text{log}}3e + {\text{log}}{e^{ - 1}} = {\text{log}}3 \\
  x + 2 = 3 \\
  x = 1,{\text{ and hence }}y = 3 \\
    \\
 $
So, the solution curve intersects this line at point (1, 3). Therefore, option (a) is correct.
As the solution curve intersects line $y = x + 2$ once, it will not cross two times. Hence, option (b) is not correct.
Case 2: Now, checking option (c), by putting $y = {(x + 2)^2}$ in eq. (2), we get,
$
  \dfrac{{{{(x + 2)}^2}}}{{x + 2}} + 2\log (x + 2) = {\text{log}}3e \\
  x + 2 + 2\log (x + 2) = {\text{log}}3 + 1 \\
  x + 2\log (x + 2) = {\text{log}}3 - 1 \\
  \log {(x + 2)^2} + \log {e^x} = \log \dfrac{3}{e} \\
  {e^x}{(x + 2)^2} = \dfrac{3}{e} \\
  {e^{x + 1}}{(x + 2)^2} = 3 \\
 $
With this expression, we will not get an exact value for $x$ and in turn, no value for $y.$ Hence, the curve is not intersecting it. Hence option (c) is not correct.
Case 3: Now, checking option (d), by putting $y = {(x + 3)^2}$ in eq. (2), we will not get any point which the solution curve touches or crosses. Hence, this option is correct.

Thus, the answers to the question are option (a) and (d).

Note: In such problems, it is noted that we shall have to check all the options for their correctness. It is easy to solve these equations by variable separable methods as we can reach the solution faster. Also, one can check and get the answer by putting the point (1, 3) value in the options. If point value satisfies the equation of a particular option then that option gives the correct answer.