Question

Find the particular solution of the following differential equation; $\dfrac{dx}{dy}=1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}}$ given that y = 1 when x = 0.

Answer 1

In order to solve this problem, we need to separate the variables and then integrate with respect to respective variables. Also, we need to know some standard formulas for integrations as $\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x}$ and $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ .

Complete step by step answer:
We are given the differential equation to solve,
We differential equation says that,
$\dfrac{dx}{dy}=1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}}$
We need to separate the variables in order to solve it.
Let's factorize the right-hand side.
We get as follows,
$\begin{aligned} & \dfrac{dx}{dy}=1\left( 1+{{x}^{2}} \right)+{{y}^{2}}\left( 1+{{x}^{2}} \right) \\\ & \dfrac{dx}{dy}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right) \\\ \end{aligned}$
Now, we are in a position to separate the variables.
By separating we get,
$\begin{aligned} & \dfrac{dx}{dy}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right) \\\ & \dfrac{dx}{\left( 1+{{x}^{2}} \right)}=\left( 1+{{y}^{2}} \right)dy \\\ \end{aligned}$
For integration, we need to know some of the standard formulas as
$\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x}$ and $\int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}}$ .
Integrating on both sides with respect to respectable variables we get,
$\begin{aligned} & \int{\dfrac{dx}{\left( 1+{{x}^{2}} \right)}}=\int{\left( 1+{{y}^{2}} \right)dy} \\\ & {{\tan }^{-1}}x=\int{dy+\int{{{y}^{2}}dy}} \\\ & {{\tan }^{-1}}x=y+\dfrac{{{y}^{3}}}{3}+C \\\ \end{aligned}$
Now, we are given the boundary conditions to eliminate C.
Hence, substituting the boundary conditions as y = 1 when x = 0, we get,
${{\tan }^{-1}}0=1+\dfrac{{{1}^{3}}}{3}+C$
Solving this for the value of C we get,
$\begin{aligned} & 0=1+\dfrac{1}{3}+C \\\ & C=\dfrac{-4}{3} \\\ \end{aligned}$
Substituting the value of C in the solution we get the final answer as follows
${{\tan }^{-1}}x=y+\dfrac{{{y}^{3}}}{3}-\dfrac{4}{3}$
Hence, the solution of the differential equation is ${{\tan }^{-1}}x=y+\dfrac{{{y}^{3}}}{3}-\dfrac{4}{3}$ .

Note:
In this problem, we need to substitute the boundary condition. The common mistake done is the value of C is not usually found. So, we need to substitute the boundary condition in order to find the value of C.