Question

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

Answer 1

To solve this differential equation, we will manipulate the given equation such that all the terms in variable x are with dx and all the terms with variable y are with dy. Then we will apply integration on both sides and carry out indeterminate differentiation. Then we substitute the values of x and y given in the question to find the value of the constant. Once we get the value of constant, we can substitute it in the integrated equation and manipulate the equation to be in standard form.

Complete step by step answer:
The equation given to us is $x\left( 1+{{y}^{2}} \right)dx-y\left( 1+{{x}^{2}} \right)dy=0$
We will take the terms $y\left( 1+{{x}^{2}} \right)dy$ on the other side of the equal to sign.
The modified equation is $x\left( 1+{{y}^{2}} \right)dx=y\left( 1+{{x}^{2}} \right)dy$
We will now perform cross multiplication to have terms of x with dx and terms of y with dy.
The equation thus becomes as $\dfrac{x}{\left( 1+{{x}^{2}} \right)}dx=\dfrac{y}{\left( 1+{{y}^{2}} \right)}dy$
Now, we have to apply integration on both sides.
$\Rightarrow \int{\dfrac{x}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{y}{\left( 1+{{y}^{2}} \right)}dy}$
We know that $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}=\log f\left( x \right)}$ . Thus, we need to make the numerator in the form of derivative of denominator on the both sides.
Derivative of $\left( 1+{{x}^{2}} \right)$ = 2x and derivative of $\left( 1+{{y}^{2}} \right)$ = 2y.
Thus, we multiply both sides with 2.
The modified equation thus will be $\int{\dfrac{2x}{\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{2y}{\left( 1+{{y}^{2}} \right)}dy}$.
As we can see, both sides are now in the form $\dfrac{f'\left( x \right)}{f\left( x \right)}$.
Now, we shall execute the integration.
$\Rightarrow \log \left( 1+{{x}^{2}} \right)=\log \left( 1+{{y}^{2}} \right)+\log C$ , where C is a constant.
We will take $\log \left( 1+{{y}^{2}} \right)$ to the other side of equal to sign.
$\Rightarrow \log \left( 1+{{x}^{2}} \right)-\log \left( 1+{{y}^{2}} \right)=\log C$
We know that $\log a-\log b=\log \left( \dfrac{a}{b} \right)$. Thus, we apply this property on the equation.
$\Rightarrow \log \left( \dfrac{1+{{x}^{2}}}{1+{{y}^{2}}} \right)=\log C$
We can now cancel the log operator from both sides.
$\Rightarrow \dfrac{1+{{x}^{2}}}{1+{{y}^{2}}}=C$
It is given that y = 1 when x = 0. So, we substitute the values of x and y in the equation $\dfrac{1+{{x}^{2}}}{1+{{y}^{2}}}=C$.
$\begin{aligned} & \Rightarrow \dfrac{1+{{\left( 0 \right)}^{2}}}{1+{{\left( 1 \right)}^{2}}}=C \\\ & \Rightarrow C=\dfrac{1}{2} \\\ \end{aligned}$

Therefore, the equation is as follows:
$\begin{aligned} & \Rightarrow \dfrac{1+{{x}^{2}}}{1+{{y}^{2}}}=\dfrac{1}{2} \\\ & \Rightarrow 1+{{y}^{2}}=2+2{{x}^{2}} \\\ & \Rightarrow y=\sqrt{2{{x}^{2}}+1} \\\ \end{aligned}$

Note: It is always better to simplify the equation before execution of integration. We have taken the constant as log C instead of C to get a simplified value of C.