Question

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

Answer 1

Hint: We will use the variables separable form to find the general integral of the given differential equation. Then we will find the constant of integral by putting the initial values given. The general equation with the arbitrary constant replaced by a particular value is called the particular solution.

Complete step by step answer:
Let us consider the given differential equation $\dfrac{dy}{dx}=1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}}.$
Since \[{{y}^{2}}\] is common in the second and third term, we will take that out to get $\dfrac{dy}{dx}=\left( 1+{{x}^{2}} \right)+{{y}^{2}}\left( 1+{{x}^{2}} \right).$
After taking \[{{y}^{2}}\] out, we will get an equation in which $1+{{x}^{2}}$ is common. So, we will take that out to get \[\dfrac{dy}{dx}=\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right).\]
Now, in the LHS, we get two terms of $x$ alone and $y$ alone. And so, we will take the variables separable form so that we can integrate the equation, $\dfrac{dy}{\left( 1+{{y}^{2}} \right)}=\left( 1+{{x}^{2}} \right)dx.$
After separating the variables, we will integrate the whole equation to get $\int{\left( \dfrac{1}{1+{{y}^{2}}} \right)}dy=\int{\left( 1+{{x}^{2}} \right)dx}.$
In the LHS, we will open the bracket to get $\int{\dfrac{1}{1+{{y}^{2}}}dy}=\int{dx}+\int{\left( 1+{{x}^{2}} \right)dx}.$
Now, we will use some of the identities to find the above integrals, $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1} }$ and $\int{\dfrac{1}{1+{{y}^{2}}}}dy={{\tan }^{-1}}y.$
When we apply the identities, we will get the general solution of the differential equation as \[{{\tan }^{-1}}y=x+\dfrac{{{x}^{3}}}{3}+C\] where $C$ is the constant of integration.
Let us apply the initial values $y=1$ when $x=0$ in the general solution to find the particular value of the constant of integration ${{\tan }^{-1}}1=0+0+C.$
So, we will get the value of the constant as $C={{\tan }^{-1}}1=\dfrac{\pi }{4},$ because we have \[{{\tan }^{-1}}\dfrac{\pi }{4}=1.\]

Hence, the particular solution is ${{\tan }^{-1}}y=x+\dfrac{{{x}^{3}}}{3}+\dfrac{\pi }{4}.$

Note: The general solution of a differential equation contains arbitrary constants. When the values of the arbitrary constants are substituted in the equation, we will get the particular solution. The solution that cannot be derived from the general solution is called the singular solution.