Question

If we have the differential equation as ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x-y}{x+y}}$ then $$$$
A.$2xy+x^2{y}^2+xy=c$ $$$$
B.$x^2+y^2-x+y=c$$$$$
C. $x^2+y^2-2xy=c$$$$$
D. $x^2-y^2-2xy=c$$$$$

Answer 1

Hint: We see that the given differential equation is a homogeneous differential equation whose standard substitution is $y=vx$.We solve the given homogeneous differential equation by putting $y=vx$ and then using the separation of variables $v,x$ to integrate. We use the standard integration $\int {\displaystyle \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}}dx=\ln f\left(x\right)$ to proceed.

Complete step-by-step solution:
We know that a differential equation consists of differentials, functions and variables. We call a first order differential equation homogeneous if ${\displaystyle \frac{dy}{dx}}=f(x,y)$ if $f(x,y)=f(kx,ky)$ for non-zero $k\in R$.We always solve homogeneous differential equation by standard substitution$y=vx$. We are given in the question the following differential equation.
$${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x-y}{x+y}}$$
Let us denote$f(x,y)={\displaystyle \frac{x-y}{x+y}}$. For any $k\in R$ we have
$$f(kx,ky)={\displaystyle \frac{kx-ky}{kx+ky}}={\displaystyle \frac{k(x-y)}{k(x+y)}}={\displaystyle \frac{x-y}{x+y}}=f(x,y)$$
So the given differential equation is a homogeneous differential equation. So let us consider$y=vx$. We differentiate both side with respect to $x$ to have;
$$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}=v{\displaystyle \frac{dx}{dx}}+x{\displaystyle \frac{dv}{dx}}\\ & \Rightarrow {\displaystyle \frac{dy}{dx}}=v+x{\displaystyle \frac{dv}{dx}}.....\left(1\right)\end{array}$$

We put $y=vx$ in the given differential equation to have;
$$\begin{array}{rl}& {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{x-vx}{x+vx}}\\ & \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1-v}{1+v}}.....\left(2\right)\end{array}$$
We equate right hand sides of (1) and (2) to have;
$$v+x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{1-v}{1+v}}$$
Now we shall use separation of variables . We have
$$\begin{array}{rl}& \Rightarrow x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{1-v}{1+v}}-v\\ & \Rightarrow x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{1-v-v(1+v)}{1+v}}\\ & \Rightarrow x{\displaystyle \frac{dv}{dx}}={\displaystyle \frac{1-2v-v^2}{1+v}}\\ & \Rightarrow {\displaystyle \frac{v+1}{1-2v-v^2}}={\displaystyle \frac{dx}{x}}\end{array}$$
We take negative sign both sides and make complete square in the left hand side to have;
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{v+1}{v^2+2v-1}}dv=-{\displaystyle \frac{dx}{x}}\\ & \Rightarrow {\displaystyle \frac{v+1}{{(v+1)}^2-2}}dv=-{\displaystyle \frac{1}{x}}dx\end{array}$$
We integrate broth dies with the respect to the corresponding variables and have ;
$$\begin{array}{rl}& \Rightarrow \int {\displaystyle \frac{v+1}{{(v+1)}^2-2}}dv=\int -{\displaystyle \frac{1}{x}}dx\\ & \Rightarrow {\displaystyle \frac{1}{2}}\int {\displaystyle \frac{2(v+1)}{{(v+1)}^2-2}}dv=-\int {\displaystyle \frac{1}{x}}dx\end{array}$$
We use the standard integration $\int {\displaystyle \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}}dx=\ln f\left(x\right)$ in both sides f the above step to have;
$$\Rightarrow {\displaystyle \frac{1}{2}}\ln |{(v+1)}^2-2|=\ln \left|x\right|+c_1$$
Let us have $c_2=\ln \left|c_1\right|$. We have
$$\begin{array}{rl}& \Rightarrow {\displaystyle \frac{1}{2}}\ln |{(v+1)}^2-2|=-\ln \left|x\right|+\ln \left|c_2\right|\\ & \Rightarrow \ln |{(v+1)}^2-2|=2\ln \left|{\displaystyle \frac{c_2}{x}}\right|\\ & \Rightarrow \ln |{(v+1)}^2-2|=\ln {\left|{\displaystyle \frac{c_2}{x}}\right|}^2\end{array}$$
We equate the arguments of respective sides to have;
$$\begin{array}{rl}& \Rightarrow {(v+1)}^2-2={\displaystyle \frac{c_2^2}{x^2}}\\ & \Rightarrow x^2(v^2-2v-1)=c_2^2\\ & \Rightarrow x^2(v^2-2v-1)=c_2^2\end{array}$$
We put back $v={\displaystyle \frac{y}{x}}$ in the above step to have;
$$\begin{array}{rl}& \Rightarrow x^2({\left({\displaystyle \frac{y}{x}}\right)}^2+2\left({\displaystyle \frac{y}{x}}\right)-1)=c_2^2\\ & \Rightarrow x^2\left({\displaystyle \frac{y^2+2xy-x^2}{x^2}}\right)=c_2^2\\ & \Rightarrow y^2+2xy-x^2=c_2^2\\ & \Rightarrow x^2-y^2-2xy=c(\text{where }c=-c_2^2)\end{array}$$
Here $c_1,c_2,c$ are real constants of integration. So the correct option is D.

Note: We note that the highest differential coefficient of a differential is called order and the highest power on the derivative when expressed in polynomial form. The given differential equation is linear because degree is 1 which makes it a linear homogeneous differential equation. We should remember the logarithmic identities $\ln \left(ab\right)=\ln a+\ln b$ and $\ln \left({\displaystyle \frac{a}{b}}\right)=\ln a-\ln b$ while solving homogeneous differential equations.