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Question

Question: A solution curve of the differential equation \(({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2}...

A solution curve of the differential equation (x2+xy+4x+2y+4)dydxy2=0,x>0,({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2} = 0,x > 0, passes through the point (1, 3). Then the solution curve
(a) intersects y=x+2 exactly at one point (b) intersects y=x+2 exactly at two points (c) intersects y=(x+2)2 (d) does NOT intersect y=(x+3)2  {\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} \\\

Explanation

Solution

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)(x,y) which is given in the question, i.e., (1, 3). Then check each option by putting the value of yy in the solution curve.

Complete step-by-step answer:
Firstly, write the expression given in the question as,
(x2+xy+4x+2y+4)dydxy2=0,x>0,({x^2} + xy + 4x + 2y + 4)\dfrac{{dy}}{{dx}} - {y^2} = 0,x > 0,
This can be rearranged in the following way,
(x2+xy+4x+2y+4)dydxy2=0 (x2+xy+4x+2y+4)dydx=y2 y2dxdy=(x2+xy+4x+2y+4)   \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 dydx\dfrac{{dy}}{{dx}} to other side of the equation, it will be dxdy\dfrac{{dx}}{{dy}})
Now, factorising the above expression, we will get
y2dxdy=(x2+4x+4)+y(x+2) y2dxdy=(x+2)2+y(x+2)  \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 y2{y^2}, we get
dxdy=(x+2y)2+(x+2y) .........(1)\Rightarrow \dfrac{{dx}}{{dy}} = {(\dfrac{{x + 2}}{y})^2} + (\dfrac{{x + 2}}{y}){\text{ }}.........{\text{(1)}}
Now, put z=(x+2y)z = (\dfrac{{x + 2}}{y}) we get
z=(x+2y) yz=x+2  z = (\dfrac{{x + 2}}{y}) \\\ yz = x + 2 \\\
And, differentiating the above equation with respect to yy, we get
z+ydzdy=dxdy z + y\dfrac{{dz}}{{dy}} = \dfrac{{dx}}{{dy}}{\text{ }}
And replacing the value of dxdy\dfrac{{dx}}{{dy}} in eq. (1), we get

z+ydzdy=z2+z ydzdy=z2 1z2dz=1ydy, now integrating, 1z2dz=1ydy  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} \\\

1z=logy+c, - \dfrac{1}{z} = \log y + c,By putting the value of zzin this equation, we will get
yx+2=logy+c- \dfrac{y}{{x + 2}} = \log y + c.Value of ccat point (1, 3) is
c=1log3 \-yx+2=logy1log3  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
yx+2+logy=1+log3 yx+2+logy=log3e ........ (2)  \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+2y = x + 2 in eq. (2), we get a unique value for xx which is
x+2x+2+log(x+2)=log3e 1+log(x+2)=log3e log(x+2)=log3e1=log3e+loge1=log3 x+2=3 x=1, and hence y=3   \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+2y = 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)2y = {(x + 2)^2} in eq. (2), we get,
(x+2)2x+2+2log(x+2)=log3e x+2+2log(x+2)=log3+1 x+2log(x+2)=log31 log(x+2)2+logex=log3e ex(x+2)2=3e ex+1(x+2)2=3  \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 xx and in turn, no value for y.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)2y = {(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.