
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
511.8k+ views
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.
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.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Father of Indian ecology is a Prof R Misra b GS Puri class 12 biology CBSE

Enzymes with heme as prosthetic group are a Catalase class 12 biology CBSE

An example of ex situ conservation is a Sacred grove class 12 biology CBSE

An orchid growing as an epiphyte on a mango tree is class 12 biology CBSE

What are the factors that influence the distribution class 12 social science CBSE

Identify the correct statement regarding cardiac activity class 12 biology CBSE
