Question

Solve $\dfrac{{dy}}{{dx}} = 1 + {x^2} + {y^2} + {x^2}{y^2}$ given that $y = 1$ and $x = 0$

Answer 1

Here we will first simplify the terms and then we will separate the unlike variables. Then we will integrate the terms using integration formulas. Then we will find the value of the constant terms by putting the given value of the variables in the equation obtained after integration. We will then back substitute the value of constant in the obtained equation to get the required solution.

Complete step by step solution:
Here we need to solve the given differential equation.
$\dfrac{{dy}}{{dx}} = 1 + {x^2} + {y^2} + {x^2}{y^2}$
Now, we will first simplify the terms by taking common terms.
$\Rightarrow \dfrac{{dy}}{{dx}} = \left( {1 + {x^2}} \right) + {y^2}\left( {1 + {x^2}} \right)$
We can see that both the terms i.e. the first and the second term have the common factor $1 + {x^2}$. So we will take this factor from both the terms.
$\Rightarrow \dfrac{{dy}}{{dx}} = \left( {1 + {x^2}} \right)\left( {1 + {y^2}} \right)$
Now, we will separate the variables on both sides. Therefore, we get
$\Rightarrow \dfrac{1}{{\left( {1 + {y^2}} \right)}}dy = \left( {1 + {x^2}} \right)dx$
Now, we will integrate both sides. So,
$\Rightarrow \int {\dfrac{1}{{\left( {1 + {y^2}} \right)}}dy} = \int {\left( {1 + {x^2}} \right)dx}$
On integrating the terms using the basic formulas ${\tan ^{ - 1}}y = \int {\dfrac{1}{{\left( {1 + {y^2}} \right)}}dy}$, $\int {kdx = kx}$ and $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ we get
$\Rightarrow {\tan ^{ - 1}}y = x + \dfrac{{{x^3}}}{3} + c$ ………….. $\left( 1 \right)$
Now, we will find the value of the constant term by substituting the values $y = 1$ and $x = 0$ in the above equation. Therefore, we get
$\Rightarrow {\tan ^{ - 1}}1 = 0 + \dfrac{{{0^3}}}{3} + c$
We know from the basic formula that ${\tan ^{ - 1}}1 = \dfrac{\pi }{4}$ .
Now, we will substitute this value in the above equation. So,
$\Rightarrow \dfrac{\pi }{4} = c$
Now, we will back substitute the value of the constant term in equation $\left( 1 \right)$. Therefore,
${\tan ^{ - 1}}y = x + \dfrac{{{x^3}}}{3} + \dfrac{\pi }{4}$
On further simplifying the terms, we get

$\Rightarrow y = \tan \left( {x + \dfrac{{{x^3}}}{3} + \dfrac{\pi }{4}} \right)$
This is the required answer.

Note:
Here we have solved the differential equation. A differential equation is defined as the equation which contains one or more functions along with their derivatives. The derivatives of any function define the rate of change of that function at a point. Integration is an inverse of differentiation and is also called anti derivative. Here, we can make a mistake by directly substituting the values of $x$ and $y$ without even solving the equation. This will give us the wrong answer.