Question

Find the particular solution of the following differential equation: $xy\dfrac{dy}{dx}=\left( x+2 \right)\left( y+2 \right);y=-1$ when $x=1$.

Answer 1

Hint: We are given a differential equation as: $xy\dfrac{dy}{dx}=\left( x+2 \right)\left( y+2 \right);y=-1$ . Firstly, write all the y-terms on one side and all the x-terms on one side. Since, we need to find a particular solution, integrate the equation. Then, substitute the values $y=-1$ and $x=1$ in the equation and get the value of the constant term. Put the constant term in the equation. Hence, we get the particular solution of the given differential equation.

Complete step-by-step solution:
Since we are given the differential equation as: $xy\dfrac{dy}{dx}=\left( x+2 \right)\left( y+2 \right)............(1)$
We need to separate the y-terms and x-terms on one side first. By separating the y-terms on one side and x-terms on one side, we can write equation (1) as:
$\dfrac{y}{\left( y+2 \right)}dy=\dfrac{\left( x+2 \right)}{x}dx............(2)$
Now, integrate equation (2) on both sides, we get:
\[\begin{align}
  & \int{\dfrac{y}{\left( y+2 \right)}dy}=\int{\dfrac{\left( x+2 \right)}{x}dx} \\
 & \int{\dfrac{y+2-2}{\left( y+2 \right)}dy}=\int{\dfrac{x}{x}dx}+\int{\dfrac{2}{x}dx} \\
 & \int{\dfrac{y+2}{\left( y+2 \right)}dy-\int{\dfrac{2}{\left( y+2 \right)}dy}}=\int{dx}+\int{\dfrac{2}{x}dx} \\
 & \int{dy-\int{\dfrac{2}{\left( y+2 \right)}dy}}=\int{dx}+\int{\dfrac{2}{x}dx}......(2)
\end{align}\]
Since, $\left[ \int{dx=x+C;\int{\dfrac{dx}{ax+b}=\log \left| ax+b \right|+C}} \right]$
We can write equation (2) as:
$y-2\log \left| y+2 \right|=x+2\log \left| x \right|+C.............(3)$
Now, we need to find the value of constant term. Put $y=-1$ and $x=1$ in the equation (3), we get:
$\begin{align}
  & \left( -1 \right)-2\log \left| -1+2 \right|=1+2\log \left| 1 \right|+C \\
 & -1-2\log \left| 1 \right|=1+2\log \left| 1 \right|+C \\
 & -2-4\log \left| 1 \right|=C……..….....(4)
\end{align}$
Since, $\log \left| 1 \right|=0$, we get:
$\begin{align}
  & C=-2-0 \\
 & =-2………......(5)
\end{align}$
Substitute the value of constant term in equation (3), we get:
$y-2\log \left| y+2 \right|=x+2\log \left| x \right|-2$
Hence, $y-2\log \left| y+2 \right|=x+2\log \left| x \right|-2$ is the particular solution of differential equation $xy\dfrac{dy}{dx}=\left( x+2 \right)\left( y+2 \right)$ when $y=-1$ and $x=1$.

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.