Question
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)+(2xy−coty)dxdy=0. Find the integrating factor.
Solution
In ordinary linear differential equation of the form dydx+P(y)x=Q(y) the integrating factor is given by e∫P(y)dy. Hence to find integrating factor we will first write the given equation in the form dydx+P(y)x=Q(y) and then find it by the formula e∫P(y)dy
Complete step-by-step answer:
Now consider the given differential equation (1+y2)+(2xy−coty)dxdy=0
Now we can rewrite the equation as
(2xy−coty)dxdy=(1+y2)
(1+y2)coty−2xy=dydx
Hence we get dydx=(1+y2)coty−(1+y2)2yx
Now rearranging the terms we get
dydx+1+y22yx=1+y2coty
Now this equation is in general form dydx+P(y)x=Q(y)
Hence we know that the integral factor of equation dydx+P(y)x=Q(y) is given by e∫P(y)dy
Now comparing the equation dydx+1+y22yx=1+y2coty with general form of linear differential equation dydx+P(y)x=Q(y) we get P(y)=1+y22y
Now we know that the integrating factor is equal to e∫P(y)dy hence first we need to find out ∫P(y)dy
Now P(y)=1+y22y
Hence we have
∫P(y)dy=∫1+y22ydy
We will solve this integration by method of substitution.
Let us substitute the denominator that is 1+y2=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
=∫tdt=lnt
Substituting the value of t we get
∫P(y)dy=ln(1+x2)
Hence we have ∫P(y)dy=ln(1+x2)
Now our integrating factor is given by e∫P(y)dy=eln(1+x2)=1+x2
Hence we value the integrating factor as 1+x2.
Note: Now we generally write any ordinary differential equation in form dxdy+P(x)y=Q(x) . Do not get confused with the general form we used which was dydx+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 dydx+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 dxdy+P(x)y=Q(x) we would not get P and Q dependent just on x but we will have y terms too.