Question

Show that the differential equation $\left( x{{e}^{\dfrac{x}{y}}}+y \right)dx=xdy$ is homogeneous. Find the particular solution of this differential equation given that $x=1$ and $y=1$?

Answer 1

We start solving the problem by recalling the definition of homogeneous function. We substitute $\left( mx,my \right)$ in place of $\left( x,y \right)$ in the differential equation to prove the homogeneity. We substitute $y=vx$ in the differential equation to solve it. We convert the differential equation in terms of v and x to find the general solution of the differential equation. We substitute the given values in the general solution to find the desired particular solution.

Complete step-by-step answer :
According to the problem, we have a differential equation $\left( x{{e}^{\dfrac{x}{y}}}+y \right)dx=xdy$. We need to prove that the given differential equation is homogeneous and we need to find the particular solution of this differential equation at $x=1$, $y=1$.
Let us assume $f\left( x,y \right)=\left( x{{e}^{\dfrac{x}{y}}}+y \right)dx-xdy$ ---(1).
We know that for a function $f\left( x,y \right)$ said to be homogeneous, it should satisfy the condition $f\left( mx,my \right)={{m}^{n}}.f\left( x,y \right)$. Let us verify this for the function in equation (1).
We substitute $\left( mx,my \right)$ in place of $\left( x,y \right)$ in equation (1).
$\Rightarrow f\left( mx,my \right)=\left( \left( mx \right).{{e}^{\dfrac{mx}{my}}}+my \right)d\left( mx \right)-\left( mx \right)d\left( my \right)$.
$\Rightarrow f\left( mx,my \right)=\left( \left( mx \right).{{e}^{\dfrac{x}{y}}}+my \right).mdx-\left( mx \right).mdy$
$\Rightarrow f\left( mx,my \right)={{m}^{2}}.\left( \left( x{{e}^{\dfrac{x}{y}}}+y \right)dx-xdy \right)$.
$\Rightarrow f\left( mx,my \right)={{m}^{2}}.f\left( x,y \right)$ ---(2).
So, we can see that the function $f\left( x,y \right)$ satisfies the condition $f\left( mx,my \right)={{m}^{n}}.f\left( x,y \right)$ of being homogeneous. So, the differential equation $\left( x{{e}^{\dfrac{x}{y}}}+y \right)dx=xdy$ is homogeneous.
Now, we find the general solution for the given differential equation $\left( x{{e}^{\dfrac{x}{y}}}+y \right)dx=xdy$ to find the particular solution at $x=1$, $y=\dfrac{\pi }{2}$.
$\Rightarrow \left( x{{e}^{\dfrac{x}{y}}}+y \right)dx=xdy$.
$\Rightarrow \dfrac{\left( x{{e}^{\dfrac{x}{y}}}+y \right)}{x}=\dfrac{dy}{dx}$.
$\Rightarrow \dfrac{x{{e}^{\dfrac{x}{y}}}}{x}+\dfrac{y}{x}=\dfrac{dy}{dx}$.
$\Rightarrow {{e}^{\dfrac{x}{y}}}+\dfrac{y}{x}=\dfrac{dy}{dx}$ ---(3).
We know that the homogeneous differential equations of the form $\dfrac{dy}{dx}=f\left( \dfrac{y}{x} \right)$ are solved by substituting $y=vx$ ---(4).
We have $y=vx$. So, we get $v=\dfrac{y}{x}$.
We differentiate with respect to ‘x’ on both sides,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( vx \right)$.
We know that the differentiation of the function of form ‘uv’ is performed as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+\dfrac{du}{dx}v$.
$\Rightarrow \dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx}$.
We know that $\dfrac{dx}{dx}=1$.
$\Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$ ---(5).
We substitute equations (4) and (5) in equation (3).
$\Rightarrow {{e}^{v}}+v=v+x\dfrac{dv}{dx}$.
$\Rightarrow {{e}^{v}}=-v+v+x\dfrac{dv}{dx}$.
$\Rightarrow {{e}^{v}}=x\dfrac{dv}{dx}$.
$\Rightarrow {{e}^{-v}}dv=\dfrac{dx}{x}$.
We perform integration on both sides.
$\Rightarrow \int{{{e}^{-v}}dv}=\int{\dfrac{dx}{x}}$.
We know that $\int{{{e}^{ax}}dx=\dfrac{{{e}^{ax}}}{a}+C}$ and $\int{\dfrac{dx}{x}=\log x+C}$.
$\Rightarrow \dfrac{{{e}^{-v}}}{-1}=\log x+C$.
$\Rightarrow -{{e}^{-v}}=\log x+C$ ---(6).
We have $y=vx$ and now we find the function v.
$\Rightarrow v=\dfrac{y}{x}$ --- (7).
We substitute equation (7) in equation (6).
$\Rightarrow -{{e}^{-\dfrac{y}{x}}}=\log x+C$ ---(8).
Let us substitute $x=1$, $y=1$ in equation (8).
$\Rightarrow -{{e}^{-\dfrac{1}{1}}}=\log 1+C$.
$\Rightarrow -{{e}^{-1}}=0+C$.
$\Rightarrow -{{e}^{-1}}=C$.
Let us substitute the value of C in equation (8).
We get the particular solution as $\Rightarrow -{{e}^{-\dfrac{y}{x}}}=\log x-{{e}^{-1}}$.
$\Rightarrow {{e}^{-1}}-{{e}^{-\dfrac{y}{x}}}=\log x$.
The particular solution at $x=1$, $y=1$ is ${{e}^{-1}}-{{e}^{-\dfrac{y}{x}}}=\log x$.
∴ The given differential equation is homogeneous and the particular solution at at $x=1$, $y=1$ is ${{e}^{-1}}-{{e}^{-\dfrac{y}{x}}}=\log x$.

Note : We should not say general solution without substituting the value of v. The particular solutions are found only if the values of variables ‘x’ and ‘y’ are given in the problem. We can’t just assume the value of arbitrary constant C without them. We use the process of substituting $y=vx$ only if the differential equation is proved as homogeneous.