Question

Solve the differential equation:
$\dfrac{dy}{dx}={{e}^{x-y}}\left( {{e}^{x}}-{{e}^{y}} \right)$

Answer 1

Here, to solve this question first we will examine what type of differential equation is it. As we can easily find that given differential equation is linear which is of form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$, we will use the concept of linear differential equation in which we need to evaluate Integrating factor by integrating P(X) and using concept of integration which is $\int{\dfrac{1}{{{x}^{n}}}dx=\dfrac{{{x}^{1-n}}}{1-n}}+C$ and after that we can proceed according to the required steps to arrive at the solution.

Complete step-by-step answer:
A first order linear differential equation is one that can be reduced to a general form:
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$
Here, P(x) and Q(x) are continuous functions in the domain of validity of the differential equation.
The solution of a linear differential equation is given as:
$I.F.\times y=\int_{{}}^{{}}{Q\left( x \right)\times I.F.dx}$
Here, I.F is the integrating factor, which is given as:
$I.F.={{e}^{\int_{{}}^{{}}{P\left( x \right)dx}}}$
Now, here the equation given to us is $\dfrac{dy}{dx}={{e}^{x-y}}\left( {{e}^{x}}-{{e}^{y}} \right)$.
We can also write it as:
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{{{e}^{y}}}\left( {{e}^{x}}-{{e}^{y}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}.{{e}^{y}}={{e}^{x}}.{{e}^{x}}-{{e}^{x}}.{{e}^{y}} \\\ & \Rightarrow \dfrac{dy}{dx}.{{e}^{y}}={{e}^{2x}}-{{e}^{x}}.{{e}^{y}} \\\ & \Rightarrow {{e}^{y}}\dfrac{dy}{dx}+{{e}^{y}}.{{e}^{x}}=2{{e}^{x}}...........\left( 1 \right) \\\ \end{aligned}$
Let us consider that ${{e}^{y}}=u$, then on differentiating both sides with respect to x, we get:
${{e}^{y}}\dfrac{dy}{dx}=\dfrac{du}{dx}$
By substituting this value in equation (1), we can write it as:
$\dfrac{du}{dx}+u.{{e}^{x}}=2{{e}^{x}}..........\left( 2 \right)$
Thus, it is a linear differential equation, so the integrating factor here is:
$I.F.={{e}^{\int_{{}}^{{}}{{{e}^{x}}.dx}}}={{e}^{{{e}^{x}}}}$
So, the solution of the equation (2) can be given as:
$u\times {{e}^{{{e}^{x}}}}=\int_{{}}^{{}}{{{e}^{2x}}\times {{e}^{{{e}^{x}}}}.dx}.............\left( 3 \right)$
Let us consider that ${{e}^{x}}=t$, so we have ${{e}^{x}}dx=dt$.
Substituting this value in equation (3), we can write eq. (3) as:
$\begin{aligned} & u\times {{e}^{t}}=\int_{{}}^{{}}{{{e}^{x}}{{e}^{x}}{{e}^{t}}.dx} \\\ & \Rightarrow u\times {{e}^{t}}=\int_{{}}^{{}}{{{e}^{x}}.{{e}^{t}}.dt} \\\ & \Rightarrow u\times {{e}^{t}}=\int_{{}}^{{}}{t{{e}^{t}}.dt}...........\left( 4 \right) \\\ \end{aligned}$
Now, we can find $\int_{{}}^{{}}{t.{{e}^{t}}.dt}$ by using integration by parts.
ILATE stands for
I - Inverse function,
L - Logarithmic function,
A - Algebraic function,
T - Trigonometric function,
E - Exponential function.
Now, if we have $\int{u(x)v(x)dx}$ , then priority of selection of function u ( x ) decreases from Inverse function to Exponential and value of integration is $\int{u(x)v(x)dx}=u(x)\int{v(x)dx-\int{\left( \dfrac{d}{dx}u(x)\int{v(x)dx} \right)}}dx$
Using ILATE rule, we can take u = t and v = ${{e}^{t}}$, so:
$\begin{aligned} & \int_{{}}^{{}}{{{e}^{t}}.t.dt}=t.\int_{{}}^{{}}{{{e}^{t}}.dt}-\int_{{}}^{{}}{\dfrac{d\left( t \right)}{dt}.\int_{{}}^{{}}{{{e}^{t}}.dt}} \\\ & \Rightarrow \int_{{}}^{{}}{{{e}^{t}}.t.dt}=t.{{e}^{t}}-\int_{{}}^{{}}{1.{{e}^{t}}.dt} \\\ & \Rightarrow \int_{{}}^{{}}{{{e}^{t}}.t.dt}=t.{{e}^{t}}-{{e}^{t}} \\\ & \Rightarrow \int_{{}}^{{}}{{{e}^{t}}.t.dt}={{e}^{t}}\left( t-1 \right) \\\ \end{aligned}$
On substituting this value in equation (4), we get:
$u\times {{e}^{t}}={{e}^{t}}\left( t-1 \right)+c$, here c is integrating constant.
Again putting $t={{e}^{x}}$ and $u={{e}^{y}}$, we get:
$\begin{aligned} & {{e}^{y}}\times {{e}^{x}}={{e}^{{{e}^{x}}}}\left( {{e}^{x}}-1 \right)+c \\\ & \Rightarrow {{e}^{y}}=\dfrac{{{e}^{{{e}^{x}}}}.{{e}^{x}}-{{e}^{{{e}^{x}}}}}{{{e}^{{{e}^{x}}}}}+c \\\ \end{aligned}$
Hence, this is the required solution.

Note: Students should note here that while choosing u and v for integrating $t.{{e}^{t}}$ , we follow the ILATE rule which is $\int{u(x)v(x)dx}=u(x)\int{v(x)dx-\int{\left( \dfrac{d}{dx}u(x)\int{v(x)dx} \right)}}dx$. Here, we have taken u = t and v = ${{e}^{t}}$, because it will diminish the repeated integral after few steps and priority of choosing of function decreases from Inverse function, Logarithmic function, Algebraic function, Trigonometric function and Exponential function and always remember the order for ILATE. Here ‘t’ is an algebraic function and ${{e}^{t}}$ is an exponential function. That is why, first preference is given to t. Students must remember general equation of Linear differential equation which is $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$, the formula for finding Integrating factor and the solution of a linear differential equation which are $I.F.\times y=\int_{{}}^{{}}{Q\left( x \right)\times I.F.dx}$ and $y.(I.F)=\int_{{}}^{{}}{(I.F)Q(x)dx}+C$ respectively. Try not to make any calculation mistakes.