Question

Show that the differential equation $2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$ is homogeneous. Find the particular solution of this differential equation, given that x = 0, when y = 1.

Answer 1

Hint: Condition for any differential equation F(x, 4) to be homogeneous or not is given as
$F\left( \lambda x,\lambda y \right)={{\lambda }^{n}}F\left( x,y \right)$
Substitute y = vx or v = xy to solve the given differential equation. Put values of (x, y) as (0, 1) to get the particular solution.

Complete step-by-step answer:
Here, given differential equation is
$2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$ …………….. (i)
Now, we need to prove this equation as a homogeneous and have to get a particular solution at x = 0 and y = 1.
We know that any differential equation will be a homogeneous one if it satisfies the condition:
$F\left( \lambda x,\lambda y \right)={{\lambda }^{n}}F\left( x,y \right)$ ……………….. (ii)
Where n is any integer.
Now, we have F(x, y) from equation (i) as
$F\left( x,y \right)=2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy=0$
And hence$F\left( \lambda x,\lambda y \right)$can be calculated by putting $x=\lambda x$ and $y=\lambda y$in the above equation. Hence, we get
\[F\left( \lambda x,\lambda y \right)=2\lambda y{{e}^{\left( \dfrac{\lambda x}{\lambda y} \right)}}dx+\left( \lambda y-2\lambda x{{e}^{\dfrac{\lambda x}{\lambda y}}} \right)dy\]
On simplifying the above equation, we get
$\begin{align}
  & F\left( \lambda x,\lambda y \right)=\lambda \left[ 2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy \right] \\
 & \Rightarrow F\left( \lambda x,\lambda y \right)={{\lambda }^{1}}F\left( x,y \right) \\
\end{align}$
Hence, we observe that the above relation is following the condition expressed in equation (ii), hence, we get to know that the given differential equation is a homogenous one. Now, let us solve the given differential equation in the following way.
Equation (i) can be divided by ‘dx’, So, we get
$\begin{align}
  & \dfrac{2y{{e}^{\dfrac{x}{y}}}dx+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)dy}{dx}=0 \\
 & \Rightarrow 2y{{e}^{\dfrac{x}{y}}}+\left( y-2x{{e}^{\dfrac{x}{y}}} \right)\dfrac{dy}{dx}=0 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-2y{{e}^{\left( \dfrac{x}{y} \right)}}}{y-2x{{e}^{\left( \dfrac{x}{y} \right)}}}..........(iii) \\
\end{align}$
Now, put y = vx in the equation (iii), since, y = vx
Now, differentiate it with respect to ‘x’ o get the value of $\dfrac{dy}{dx}$. Hence, we get
$\dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx}$
Where, we have applied relationship:
$\dfrac{dy}{dx}\left( u.v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ where u and v are general functions in multiplication.
Hence, we get
$\dfrac{dy}{dx}=v+x\dfrac{du}{dx}$……………… (iv)
Now, put y =vx in equation (iii) and replace $\dfrac{dy}{dx}$by $v+x\dfrac{du}{dx}$
Hence, we get
$\begin{align}
  & v+x\dfrac{du}{dx}=\dfrac{-2vx{{e}^{\dfrac{x}{vx}}}}{vx-2x{{e}^{\dfrac{x}{vx}}}} \\
 & \Rightarrow v+x\dfrac{du}{dx}=\dfrac{-2v{{e}^{\dfrac{1}{v}}}}{v-2{{e}^{\dfrac{1}{v}}}} \\
 & \Rightarrow x\dfrac{du}{dx}=\dfrac{-2v{{e}^{\dfrac{1}{v}}}}{v-2{{e}^{\dfrac{1}{v}}}}-\dfrac{v}{1} \\
 & \Rightarrow x\dfrac{du}{dx}=\dfrac{-2v{{e}^{\dfrac{1}{v}}}-{{v}^{2}}+2v{{e}^{\dfrac{1}{v}}}}{v-2{{e}^{\dfrac{1}{v}}}} \\
 & \Rightarrow x\dfrac{du}{dx}=\dfrac{-{{v}^{2}}}{v-2{{e}^{\dfrac{1}{v}}}} \\
\end{align}$
Now, we can separate the variables ‘v’ and ‘x’ as
$\dfrac{v-2{{e}^{\dfrac{1}{v}}}}{{{v}^{2}}}dv=-\dfrac{dx}{x}$
Now, integrate both the sides to get a solution of the differential equation.
Hence, we get
$\int{\dfrac{v-2{{e}^{\dfrac{1}{v}}}}{{{v}^{2}}}dv=-\int{\dfrac{1}{x}dx}}$ ……………. (v)
Let ${{I}_{1}}=\int{\dfrac{v-2{{e}^{\dfrac{1}{v}}}}{{{v}^{2}}}dv}$and ${{I}_{2}}=-\int{\dfrac{1}{x}dx}$
Let us solve ${{I}_{1}}$ and ${{I}_{2}}$ to get equation (v).
Hence, ${{I}_{1}}$ can be written as
$\begin{align}
  & {{I}_{1}}=\int{\dfrac{1}{v}dv}-\int{\dfrac{2{{e}^{\dfrac{1}{v}}}}{{{v}^{2}}}}dv \\
 & \\
\end{align}$
Now, we know
$\int{\dfrac{1}{x}dx}={{\log }_{e}}x$

Hence, ${{I}_{1}}$can be given as
${{I}_{1}}={{\log }_{e}}v-2\int{\dfrac{{{e}^{\dfrac{1}{v}}}}{{{v}^{2}}}dv}$ …………………… (vi)
Suppose
$\dfrac{1}{v}=t$ for second integral
Now, differentiate it with respect to ‘t’.
We know $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$
Hence,
$\begin{align}
  & \dfrac{d}{dt}\left( {{v}^{-1}} \right)=\dfrac{d}{dt}\left( t \right) \\
 & \Rightarrow \dfrac{-1}{{{v}^{2}}}\dfrac{dv}{dt}=1 \\
 & \Rightarrow \dfrac{1}{{{v}^{2}}}dv=-dt \\
\end{align}$
Now, replace $\dfrac{1}{v}$ by t and $\dfrac{1}{{{v}^{2}}}dv$ by ‘-dt’ in equation (vi) for the integral part only.
$\begin{align}
  & {{I}_{1}}={{\log }_{e}}v-2\int{\left( -{{e}^{t}} \right)dt} \\
 & \Rightarrow {{I}_{1}}={{\log }_{e}}v+2\int{{{e}^{t}}dt} \\
\end{align}$
 We know $\int{{{e}^{x}}dx}={{e}^{x}}$. Hence,
${{I}_{1}}={{\log }_{e}}v+2{{e}^{t}}+{{C}_{1}}$ where ${{C}_{1}}$ is a constant.
Put $t=\dfrac{1}{v}$ . hence, we get
${{I}_{1}}={{\log }_{e}}v+2{{e}^{\dfrac{1}{v}}}+{{C}_{1}}$…………… (vii)
Now, we can get value of ${{I}_{2}}$ by using relation $\int{\dfrac{1}{x}dx}={{\log }_{e}}x$
Hence, we get
${{I}_{2}}=\int{\dfrac{1}{x}dx}={{\log }_{e}}x+{{C}_{2}}$………….. (viii)
Where ${{C}_{2}}$ is a constant.
Now put values of ${{I}_{1}}$ and ${{I}_{2}}$from equations (vii) and (viii), we get
${{\log }_{e}}v+2{{e}^{\dfrac{1}{v}}}+{{C}_{1}}=-{{\log }_{e}}x+{{C}_{2}}$
As we supposed y = vx, hence the value of v can be given as $v=\dfrac{y}{x}$.
Hence, we get the above equation as
${{\log }_{e}}\left( \dfrac{y}{x} \right)+2{{e}^{\left( \dfrac{x}{y} \right)}}=-{{\log }_{e}}x+\left( {{c}_{2}}-{{c}_{1}} \right)$
Now replace ${{c}_{2}}-{{c}_{1}}=c$.
Hence, general solution of given differentiable be
$\begin{align}
  & {{\log }_{e}}\left( \dfrac{y}{x} \right)+2{{e}^{\left( \dfrac{x}{y} \right)}}=-{{\log }_{e}}x+C \\
 & \Rightarrow {{\log }_{e}}\left( \dfrac{y}{x} \right)+{{\log }_{e}}x+2{{e}^{\left( \dfrac{x}{y} \right)}}=C\ldots \ldots \ldots \ldots \ldots \text{ }\left( ix \right) \\
\end{align}$
Now, use logarithm identity given as ${{\log }_{c}}a+{{\log }_{c}}b={{\log }_{c}}ab$ . Hence, we get
$\begin{align}
  & {{\log }_{e}}\left( \dfrac{y}{x}\times x \right)+2{{e}^{\left( \dfrac{x}{y} \right)}}=C \\
 & \Rightarrow {{\log }_{e}}y+2{{e}^{\dfrac{x}{y}}}=C...........(x) \\
\end{align}$
Now, put x = 0 and y = 1 in the above equation to get a particular solution. Hence by putting values of ‘x’ and ‘y’, we get
${{\log }_{e}}\left( 1 \right)+2{{e}^{\dfrac{0}{1}}}=C$
Value of ${{\log }_{e}}1=0$ and ${{e}^{o}}=1$, hence, we get
0 + 2 = C
Or C = 2
Hence, particular solution can be given from equation (x) by putting C = 2, we get
${{\log }_{e}}y+2{{e}^{\left( \dfrac{x}{y} \right)}}=2$

Note: One can use x = vy to solve the given homogeneous equation and replace $\dfrac{dx}{dy}$ by $v+y\dfrac{dv}{dy}$ to get the solution.
One may not able to put x = 0 and y = 1 in equation ${{\log }_{e}}\left( \dfrac{y}{x} \right)+{{\log }_{e}}x+2{{e}^{\left( \dfrac{x}{y} \right)}}=C$ as ${{\log }_{e}}\left( \dfrac{y}{x} \right)$ and ${{\log }_{e}}\left( x \right)$will not accept x = 0. So, first simplify the equation to get the equation ${{\log }_{e}}y+2{{e}^{\dfrac{x}{y}}}=C$, then try to put x = 1 and y = 1 in it.
Calculation is the important part of the question as well, so take care with the calculating part as well.