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Question: Choose the correct option for the following question given below: Solve the following differential...

Choose the correct option for the following question given below:
Solve the following differential equation:
xdx+ydy=xdy−ydxxdx + ydy = xdy - ydx
A. ln⁡(x2+y2)=2tan⁡−1(yx)+c\ln \left( {{x^2} + {y^2}} \right) = 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + c
B. ln⁡(x+y)2=2tan⁡−1(yx)+c\ln {\left( {x + y} \right)^2} = 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + c
C. ln⁡(x2+y2)=2tan⁡−1(xy)+c\ln \left( {{x^2} + {y^2}} \right) = 2{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right) + c
D. ln⁡(x+y)2=2tan⁡−1(xy)+c\ln {\left( {x + y} \right)^2} = 2{\tan ^{ - 1}}\left( {\dfrac{x}{y}} \right) + c

Explanation

Solution

We consider writing one variable in the form of another variable by assigning a value to a differentiable constant, that is vv. We then arrange the terms such that all the one type variable terms are on one side and integrate it with respect to the present variable. Arranging the terms and simplify the solution until we get the most simplified answer.

Complete step-by-step solution:
Given differential equation;
xdx+ydy=xdy−ydxxdx + ydy = xdy - ydx
Dividing both the sides by dxdx, we get;
⇒x+ydydx=xdydx−y−−−−(1)\Rightarrow x + y\dfrac{{dy}}{{dx}} = x\dfrac{{dy}}{{dx}} - y - - - - \left( 1 \right)
Now, let us consider taking yx\dfrac{y}{x} in terms of a variable vv
v=yxv = \dfrac{y}{x}
⇒y=vx\Rightarrow y = vx
Differentiating both sides with respect to dxdx, we get;
⇒dydx=v+xdvdx\Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}
Substituting the dydx\dfrac{{dy}}{{dx}} and yy in the equation 11, we get;
⇒x+xv(xdvdx+v)=x(xdvdx+v)−xv\Rightarrow x + xv\left( {x\dfrac{{dv}}{{dx}} + v} \right) = x\left( {x\dfrac{{dv}}{{dx}} + v} \right) - xv
Simplifying the equation, we get;
⇒x+xv(xdvdx+v)=x2dvdx+xv−xv\Rightarrow x + xv\left( {x\dfrac{{dv}}{{dx}} + v} \right) = {x^2}\dfrac{{dv}}{{dx}} + xv - xv
Reducing the equation, we get;
⇒x+xv(xdvdx+v)=x2dvdx\Rightarrow x + xv\left( {x\dfrac{{dv}}{{dx}} + v} \right) = {x^2}\dfrac{{dv}}{{dx}}
Expanding the equation, we get;
⇒x+x2vdvdx+xv2=x2dvdx\Rightarrow x + {x^2}v\dfrac{{dv}}{{dx}} + x{v^2} = {x^2}\dfrac{{dv}}{{dx}}
Rearranging the terms, we get;
⇒x+xv2=x2dvdx−x2vdvdx\Rightarrow x + x{v^2} = {x^2}\dfrac{{dv}}{{dx}} - {x^2}v\dfrac{{dv}}{{dx}}
Taking the common terms out, we get;
⇒x(1+v2)=dvdxx2(1−v)\Rightarrow x\left( {1 + {v^2}} \right) = \dfrac{{dv}}{{dx}}{x^2}\left( {1 - v} \right)
Cancelling the common terms out, we get;
⇒(1+v2)=dvdxx(1−v)\Rightarrow \left( {1 + {v^2}} \right) = \dfrac{{dv}}{{dx}}x\left( {1 - v} \right)
Re-arranging the terms, we get;
⇒dvdx=1+v2x(1−v)\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{1 + {v^2}}}{{x\left( {1 - v} \right)}}
Taking the negative sign from the denominator as the common term, we get;
⇒dvdx=−1−v2x(v−1)\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{ - 1 - {v^2}}}{{x\left( {v - 1} \right)}}
Now, we rearrange the terms such that the xxterms are on one side.
⇒dvdx(v−1−v2−1)=1x\Rightarrow \dfrac{{dv}}{{dx}}\left( {\dfrac{{v - 1}}{{ - {v^2} - 1}}} \right) = \dfrac{1}{x}
We can write the above equation as,
⇒(v−1−v2−1)dvdx=1x\Rightarrow \left( {\dfrac{{v - 1}}{{ - {v^2} - 1}}} \right)\dfrac{{dv}}{{dx}} = \dfrac{1}{x}
Integrating both the sides with respect to xx, we get;
∫v−1−v2−1dvdxdx=∫1xdx\int {\dfrac{{v - 1}}{{ - {v^2} - 1}}} \dfrac{{dv}}{{dx}}dx = \int {\dfrac{1}{x}} dx
⇒tan⁡−1v−12log⁡(v2+1)=log⁡x+c\Rightarrow {\tan ^{ - 1}}v - \dfrac{1}{2}\log \left( {{v^2} + 1} \right) = \log x + c
There is a constant added because it is integration.
Now substituting the value of vv in terms of yy, we get;
⇒tan⁡−1(yx)−12log⁡((yx)2+1)=log⁡x+c\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) - \dfrac{1}{2}\log \left( {{{\left( {\dfrac{y}{x}} \right)}^2} + 1} \right) = \log x + c
Rearranging the terms, we get;
⇒tan⁡−1(yx)=log⁡x+12log⁡((yx)2+1)+c\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = \log x + \dfrac{1}{2}\log \left( {{{\left( {\dfrac{y}{x}} \right)}^2} + 1} \right) + c
By taking the L.C.M. of the denominator and taking it to the other side, we get;
⇒2tan⁡−1(yx)=2log⁡x+log⁡((yx)2+1)+c\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = 2\log x + \log \left( {{{\left( {\dfrac{y}{x}} \right)}^2} + 1} \right) + c
By applying the formula of logarithmic powers, we get;
⇒2tan⁡−1(yx)=log⁡x2+log⁡((yx)2+1)+c\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = \log {x^2} + \log \left( {{{\left( {\dfrac{y}{x}} \right)}^2} + 1} \right) + c
By using the logarithmic theorem of addition, we get;
⇒2tan⁡−1(yx)=log⁡(x2(y2x2+1))+c\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = \log \left( {{x^2}\left( {\dfrac{{{y^2}}}{{{x^2}}} + 1} \right)} \right) + c
By multiplying the terms, we get;
⇒2tan⁡−1(yx)=log⁡(x2y2x2+x2)+c\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = \log \left( {{x^2}\dfrac{{{y^2}}}{{{x^2}}} + {x^2}} \right) + c
By cancelling the common terms, we have;
⇒2tan⁡−1(yx)=log⁡(x2+y2)+c\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) = \log \left( {{x^2} + {y^2}} \right) + c
Rearranging the terms, we get;
⇒log⁡(x2+y2)=2tan⁡−1(yx)+c\Rightarrow \log \left( {{x^2} + {y^2}} \right) = 2{\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) + c

∴\therefore The correct option is A.

Note: A differential equation is an equation that relates one or more functions and their derivatives. Mainly the study of differential equations consists of the study of their solutions (the set of functions that satisfy each equation), and of the properties of their solutions. Only one simplest differential equation is solvable by explicit formulas, however, many properties of solution of a given differential equation may be determined without computing them exactly.