Question

In the given equation $(1+{{y}^{2}})+(2xy-\cot y)\dfrac{dy}{dx}=0$. Find the integrating factor.

Answer 1

In ordinary linear differential equation of the form $\dfrac{dx}{dy}+P(y)x=Q(y)$ the integrating factor is given by ${{e}^{\int{P(y)dy}}}$. Hence to find integrating factor we will first write the given equation in the form $\dfrac{dx}{dy}+P(y)x=Q(y)$ and then find it by the formula ${{e}^{\int{P(y)dy}}}$

Complete step-by-step answer:
Now consider the given differential equation $(1+{{y}^{2}})+(2xy-\cot y)\dfrac{dy}{dx}=0$
Now we can rewrite the equation as
$(2xy-\cot y)\dfrac{dy}{dx}=(1+{{y}^{2}})$
$\dfrac{\cot y-2xy}{(1+{{y}^{2}})}=\dfrac{dx}{dy}$
Hence we get $\dfrac{dx}{dy}=\dfrac{\cot y}{(1+{{y}^{2}})}-\dfrac{2yx}{(1+{{y}^{2}})}$
Now rearranging the terms we get
$\dfrac{dx}{dy}+\dfrac{2y}{1+{{y}^{2}}}x=\dfrac{\cot y}{1+{{y}^{2}}}$
Now this equation is in general form $\dfrac{dx}{dy}+P(y)x=Q(y)$
Hence we know that the integral factor of equation $\dfrac{dx}{dy}+P(y)x=Q(y)$ is given by ${{e}^{\int{P(y)dy}}}$
Now comparing the equation $\dfrac{dx}{dy}+\dfrac{2y}{1+{{y}^{2}}}x=\dfrac{\cot y}{1+{{y}^{2}}}$ with general form of linear differential equation $\dfrac{dx}{dy}+P(y)x=Q(y)$ we get $P(y)=\dfrac{2y}{1+{{y}^{2}}}$
Now we know that the integrating factor is equal to ${{e}^{\int{P(y)dy}}}$ hence first we need to find out $\int{P(y)dy}$
Now $P(y)=\dfrac{2y}{1+{{y}^{2}}}$
Hence we have
$\int{P(y)dy}=\int{\dfrac{2y}{1+{{y}^{2}}}dy}$
We will solve this integration by method of substitution.
Let us substitute the denominator that is $1+{{y}^{2}}=t$ then differentiating on both sides we get $2ydy=dt$
Hence using this substitution our denominator becomes t and numerator becomes dt hence we get
$=\int{\dfrac{dt}{t}=\ln t}$
Substituting the value of t we get
$\int{P(y)dy}=\ln (1+{{x}^{2}})$
Hence we have $\int{P(y)dy}=\ln (1+{{x}^{2}})$
Now our integrating factor is given by ${{e}^{\int{P(y)dy}}}={{e}^{\ln (1+{{x}^{2}})}}=1+{{x}^{2}}$
Hence we value the integrating factor as $1+{{x}^{2}}$.

Note: Now we generally write any ordinary differential equation in form $\dfrac{dy}{dx}+P(x)y=Q(x)$ . Do not get confused with the general form we used which was $\dfrac{dx}{dy}+P(y)x=Q(y)$. This is just a change of variables. Now in our question y is the independent variable and x is the dependent variable so we write the general form as $\dfrac{dx}{dy}+P(y)x=Q(y)$ .
As we know the functions P and Q are just functions of the dependent variable which in our case is y. Now if we try to write the equation in the form $\dfrac{dy}{dx}+P(x)y=Q(x)$ we would not get P and Q dependent just on x but we will have y terms too.