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

In the given equation (1+y2)+(2xycoty)dydx=0(1+{{y}^{2}})+(2xy-\cot y)\dfrac{dy}{dx}=0. Find the integrating factor.

Explanation

Solution

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

Complete step-by-step answer:
Now consider the given differential equation (1+y2)+(2xycoty)dydx=0(1+{{y}^{2}})+(2xy-\cot y)\dfrac{dy}{dx}=0
Now we can rewrite the equation as
(2xycoty)dydx=(1+y2)(2xy-\cot y)\dfrac{dy}{dx}=(1+{{y}^{2}})
coty2xy(1+y2)=dxdy\dfrac{\cot y-2xy}{(1+{{y}^{2}})}=\dfrac{dx}{dy}
Hence we get dxdy=coty(1+y2)2yx(1+y2)\dfrac{dx}{dy}=\dfrac{\cot y}{(1+{{y}^{2}})}-\dfrac{2yx}{(1+{{y}^{2}})}
Now rearranging the terms we get
dxdy+2y1+y2x=coty1+y2\dfrac{dx}{dy}+\dfrac{2y}{1+{{y}^{2}}}x=\dfrac{\cot y}{1+{{y}^{2}}}
Now this equation is in general form dxdy+P(y)x=Q(y)\dfrac{dx}{dy}+P(y)x=Q(y)
Hence we know that the integral factor of equation dxdy+P(y)x=Q(y)\dfrac{dx}{dy}+P(y)x=Q(y) is given by eP(y)dy{{e}^{\int{P(y)dy}}}
Now comparing the equation dxdy+2y1+y2x=coty1+y2\dfrac{dx}{dy}+\dfrac{2y}{1+{{y}^{2}}}x=\dfrac{\cot y}{1+{{y}^{2}}} with general form of linear differential equation dxdy+P(y)x=Q(y)\dfrac{dx}{dy}+P(y)x=Q(y) we get P(y)=2y1+y2P(y)=\dfrac{2y}{1+{{y}^{2}}}
Now we know that the integrating factor is equal to eP(y)dy{{e}^{\int{P(y)dy}}} hence first we need to find out P(y)dy\int{P(y)dy}
Now P(y)=2y1+y2P(y)=\dfrac{2y}{1+{{y}^{2}}}
Hence we have
P(y)dy=2y1+y2dy\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+y2=t1+{{y}^{2}}=t then differentiating on both sides we get 2ydy=dt2ydy=dt
Hence using this substitution our denominator becomes t and numerator becomes dt hence we get
=dtt=lnt=\int{\dfrac{dt}{t}=\ln t}
Substituting the value of t we get
P(y)dy=ln(1+x2)\int{P(y)dy}=\ln (1+{{x}^{2}})
Hence we have P(y)dy=ln(1+x2)\int{P(y)dy}=\ln (1+{{x}^{2}})
Now our integrating factor is given by eP(y)dy=eln(1+x2)=1+x2{{e}^{\int{P(y)dy}}}={{e}^{\ln (1+{{x}^{2}})}}=1+{{x}^{2}}
Hence we value the integrating factor as 1+x21+{{x}^{2}}.

Note: Now we generally write any ordinary differential equation in form dydx+P(x)y=Q(x)\dfrac{dy}{dx}+P(x)y=Q(x) . Do not get confused with the general form we used which was dxdy+P(y)x=Q(y)\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 dxdy+P(y)x=Q(y)\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 dydx+P(x)y=Q(x)\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.