Question

Which of the following curve satisfying the differential equation, $ \left( {{x^2} - {y^2}} \right)dx + 2xydy = 0 $ and passing through the point $ \left( {1,1} \right) $ ?
(A) A circle of radius two
(B) A circle of radius one
(C) A hyperbola
(D) An ellipse

Answer 1

Hint: For judging the curve of a differential equation you need to find a solution for the differential equation. First, rewrite the given equation with the left hand side as $ \dfrac{{dy}}{{dx}} $ and then use substitution $ y = mx $ . This will change the equation into variable separable form. Now you can integrate both sides after separating variables on both sides to find the solution equation. If you still can find the nature of the curve, try changing it in the standard form of a curve.

Complete step-by-step solution:
Here we are given a differential equation $ \left( {{x^2} - {y^2}} \right)dx + 2xydy = 0 $ and we are asked to find the curve that this equation represents. But we can only determine that with an actual equation of the curve. Hence, we need to find the solution of this differential equation that satisfies the point $ \left( {1,1} \right) $
We have the equation that can be rewritten as: $ \left( {{x^2} - {y^2}} \right)dx + 2xydy = 0 \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\left( {{x^2} - {y^2}} \right)}}{{2xy}} = \dfrac{{\left( {{y^2} - {x^2}} \right)}}{{2xy}} $
So, we get a differential equation of first order and it has the sum of power in each term as $ 2 $ for both numerator and denominator. Hence, we can use the substitution $ y = mx $ , which will give you:
 $ \Rightarrow \dfrac{{d\left( {mx} \right)}}{{dx}} = \dfrac{{{{\left( {mx} \right)}^2} - {x^2}}}{{2x \times mx}} $
This can be further solved by differentiating $ mx $ using product rule of differentiation, i.e. $ \dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right)\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] $
 $ \Rightarrow \dfrac{{d\left( {mx} \right)}}{{dx}} = \dfrac{{{{\left( {mx} \right)}^2} - {x^2}}}{{2x \times mx}} \Rightarrow m + x\dfrac{{dm}}{{dx}} = \dfrac{{{x^2}\left( {{m^2} - 1} \right)}}{{2m{x^2}}} = \dfrac{{\left( {{m^2} - 1} \right)}}{{2m}} $
Therefore, we get $ m + x\dfrac{{dm}}{{dx}} = \dfrac{{\left( {{m^2} - 1} \right)}}{{2m}} $ . Now we can shift the terms of $ m $ to left and terms of $ x $ to the right side
 $ \Rightarrow m + x\dfrac{{dm}}{{dx}} = \dfrac{{\left( {{m^2} - 1} \right)}}{{2m}} \Rightarrow x\dfrac{{dm}}{{dx}} = \dfrac{{{m^2} - 1 - 2{m^2}}}{{2m}} \Rightarrow \dfrac{{2m}}{{\left( { - 1 - {m^2}} \right)}}dm = \dfrac{{dx}}{x} $
So, we have separated the variables on both sides. Now we can easily integrate both the sides to get the solution.
 \[ \Rightarrow \int {\dfrac{{2m}}{{\left( { - 1 - {m^2}} \right)}}dm} = \int {\dfrac{{dx}}{x} \Rightarrow - \int {\dfrac{{2m}}{{1 + {m^2}}}} } dm = \int {\dfrac{{dx}}{x}} \]
Now, on the left side, we can substitute $ {m^2} = g \Rightarrow 2mdm = dg $ and for the right side we know $ \int {\dfrac{{dx}}{x}} = \ln x + C $
Let’s use these two in the previous equation
 \[ \Rightarrow - \int {\dfrac{{2m}}{{1 + {m^2}}}} dm = \int {\dfrac{{dx}}{x}} \Rightarrow - \int {\dfrac{{dg}}{{1 + g}}} = \ln \left| x \right| + C\]
Again using $ \int {\dfrac{{dx}}{x}} = \ln x + C $ , we can write it as:
 \[ \Rightarrow - \int {\dfrac{{dg}}{{1 + g}}} = \ln \left| x \right| + C \Rightarrow - \ln \left| {1 + g} \right| = \ln \left| x \right| + \ln \left| c \right| \Rightarrow \ln \left| {\dfrac{1}{{1 + g}}} \right| = \ln \left| x \right| + \ln \left| c \right|\]
Now, we can use some properties of a logarithmic function such as $ \ln {a^b} = b\ln a{\text{ and }}\ln a + \ln b = \ln ab $ and put back our substitution in the above equation
 \[ \Rightarrow \ln \left| {\dfrac{1}{{1 + g}}} \right| = \ln \left| x \right| + \ln \left| c \right| \Rightarrow \ln \left| {\dfrac{1}{{1 + {m^2}}}} \right| = \ln \left| {xc} \right|\]
Now, we can remove the logarithmic function from both sides by equating what’s inside
 $ \Rightarrow \dfrac{1}{{1 + {m^2}}} = xc $
But we know, $ m = \dfrac{y}{x} $ . Therefore, we get: $ \Rightarrow \dfrac{1}{{1 + \dfrac{{{y^2}}}{{{x^2}}}}} = \dfrac{{{x^2}}}{{{x^2} + {y^2}}} = xc $
After simplifying the above equation we get: $ {x^2} + {y^2} - \dfrac{x}{c} = 0 $
So, this is the equation of the required curve. It is also given that this curve passes from point $ \left( {1,1} \right) $
So, we can satisfy this point the curve to find the value for the constant c
 $ \Rightarrow $ For $ \left( {1,1} \right) $ : $ {x^2} + {y^2} - \dfrac{x}{c} = 0 \Rightarrow 1 + 1 - \dfrac{1}{c} = 0 \Rightarrow c = \dfrac{1}{2} $
Thus, our equation of the curve becomes: $ {x^2} + {y^2} - 2x = 0 $
The same can also be written as: $ {x^2} + {y^2} - 2x + 1 - 1 = 0 \Rightarrow {\left( {x - 1} \right)^2} + {y^2} = 1 $
As we know that the standard equation of a circle with centre at $ \left( {a,b} \right) $ and radius $ 'r' $ is $ {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2} $
So, our equation $ {\left( {x - 1} \right)^2} + {y^2} = 1 $ represents a circle with centre at $ \left( {1,0} \right) $ and with radius $ 1unit $ .

Hence, the option (B) is the correct answer

Note: Notice that the equation $ {x^2} + {y^2} - 2x = 0 $ is in its general form. It is difficult to judge the curve in its general form, try to change the equation in the standard form to figure the curve when it is not clear. We use the equation of circle but the standard equations of hyperbola is given by $ \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} - \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 $ and for ellipse it is $ \dfrac{{{{\left( {x - h} \right)}^2}}}{{{a^2}}} + \dfrac{{{{\left( {y - k} \right)}^2}}}{{{b^2}}} = 1 $ .