Question

Find the particular solution of the following differential equation:
$\left( x+1 \right)\dfrac{dy}{dx}=2{{e}^{-y}}-1;y=0\text{ and }x=0$

Answer 1

We have the following differential equation: $\left( x+1 \right)\dfrac{dy}{dx}=2{{e}^{-y}}-1$. Separate x terms on one side and y terms on the other side. Then integrate both the sides to get an equation in terms of x and y. Now, put x = 0 and y = 0 in the equation to get the value of the constant term. Then substitute the constant term in the equation to get the particular solution of the given differential equation.

Complete step-by-step answer:
We have a differential equation as:
$\left( x+1 \right)\dfrac{dy}{dx}=2{{e}^{-y}}-1......(1)$
Now, by separating the x and y terms we can write equation (1) as:
$\begin{aligned} & \dfrac{dy}{2{{e}^{-y}}-1}=\dfrac{dx}{\left( x+1 \right)} \\\ & \dfrac{{{e}^{y}}dy}{2-{{e}^{y}}}=\dfrac{dx}{\left( x+1 \right)}......(2) \\\ \end{aligned}$
Now, integrate equation (2), we get:
$\int{\dfrac{{{e}^{y}}dy}{2-{{e}^{y}}}}=\int{\dfrac{dx}{\left( x+1 \right)}}......(3)$
Let
$\begin{aligned} & 2-{{e}^{y}}=t \\\ & -{{e}^{y}}dy=dt \\\ \end{aligned}$
So, we can write:
$\int{\dfrac{{{e}^{y}}dy}{2-{{e}^{y}}}}=-\int{\dfrac{dt}{t}}$
Since, $\int{\dfrac{dx}{x}=\log \left| x \right|+C}$
So, we have: $-\int{\dfrac{dx}{t}=-\log \left| t \right|}$
Since $2-{{e}^{y}}=t$
So, we have:
$\int{\dfrac{{{e}^{y}}dy}{2-{{e}^{y}}}}=-\log \left| 2-{{e}^{y}} \right|$
Similarly, for $\int{\dfrac{dx}{\left( x+1 \right)}=\log \left| x+1 \right|+C}$
Put the values in equation (2), we get:
$\begin{aligned} & -\log \left| 2-{{e}^{y}} \right|=\log \left| x+1 \right|+C \\\ & \log \left| \dfrac{1}{2-{{e}^{y}}} \right|=\log \left| C\left( x+1 \right) \right|......(3) \\\ \end{aligned}$
$\left| \dfrac{1}{2-{{e}^{y}}} \right|=\left| C\left( x+1 \right) \right|......(4)$
Now, we have x = 0 and y = 0, so we get:

& \Rightarrow \left| \dfrac{1}{2-{{e}^{0}}} \right|=\left| C\left( 0+1 \right) \right| \\\ & \Rightarrow C=\dfrac{1}{2} \\\ \end{aligned}$$ Put the values in equation (4), we get: $$\begin{aligned} & \left| \dfrac{1}{2-{{e}^{y}}} \right|=\left| \dfrac{1}{2}\left( x+1 \right) \right| \\\ & \left( 2-{{e}^{y}} \right)=\dfrac{2}{x+1} \\\ & {{e}^{y}}=2-\dfrac{2}{x+1} \\\ \end{aligned}$$ Hence, $${{e}^{y}}=2-\dfrac{2}{x+1}$$ is the particular solution of $\left( x+1 \right)\dfrac{dy}{dx}=2{{e}^{-y}}-1$ **Note:** Always remember that while integration a given function adds a constant term in the answer. Also, after integrating, do not assume that the value of the constant term is zero unless stated in the question. First find the value of the constant term and then proceed further to find the particular solution of the differential equations. Do not directly write the integrated equation as a particular solution. That is an incomplete answer.