Question #96321
Obtain the solution for the initial value problem y'+(cotx)y=xcscx, y left frac{pi}{2} right=1
1
Expert's answer
2019-10-14T12:47:21-0400

We find the general solution of the equation


dydx+(cotx)y=xcsc(x)\frac{dy}{dx} +(cotx)y=x csc(x)


This equation is a linear equation of the form


dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x)


where

P(x)=cot(x),𝑄(π‘₯)=π‘₯csc(x)P(x) = cot (x),𝑄(π‘₯) = π‘₯ csc(x)


Integrating factor :


IF(x)=eβˆ«π‘ƒ(π‘₯)𝑑x=e∫cot(x)𝑑x==eβˆ«π‘‘(sinπ‘₯)sinx=eln∣sinπ‘₯∣+𝐢IF(x) = e^ {\int 𝑃(π‘₯)𝑑x }=e^ {\int cot (x) 𝑑x }=\\ =e^ {\int \frac{𝑑(sinπ‘₯) }{sin x} }= e^ {ln|sinπ‘₯| + 𝐢}


We are looking for a particular integrating factor, so we take C=0.


Then we get 


IF(x)=sin(π‘₯)IF (x)= sin(π‘₯)


Multiply both sides of the equation by IF(x)=sin(π‘₯)IF (x)= sin(π‘₯)


sin(π‘₯)dydx+(sin(π‘₯)β‹…cot(x))y=xcsc(x)β‹…sin(π‘₯)sin(π‘₯) \frac{dy}{dx} +(sin(π‘₯) β‹…cot(x))y=x csc(x)β‹… sin(π‘₯)

or


sin(π‘₯)dydx+(cos(π‘₯))𝑦=π‘₯sin(π‘₯) \frac{dy}{dx} + (cos(π‘₯))𝑦 = π‘₯


i.e.


ddx(𝑦⋅sinπ‘₯)=π‘₯.\frac{d}{dx}(𝑦 β‹… sinπ‘₯) = π‘₯.


Integrate both sides of this equation we get the general solution


∫ddx(𝑦⋅sinπ‘₯)dx=∫π‘₯dx⇒𝑦⋅sinπ‘₯=π‘₯22+C\int \frac{d}{dx}(𝑦 β‹… sinπ‘₯)dx = \int π‘₯ dx \Rightarrow 𝑦 β‹… sinπ‘₯ = \frac{π‘₯^2}{2} +C ,


where C - constant defined by the initial conditions. 



Now we find C and solve the initial value problem.


Substitute the initial condition


y(Ο€2)=1y(\frac{\pi}{2}) =1


into the general solution:


y(π2)⋅sin(π2)=(π2)22+C⇒1⋅sin(π2)=12(π2)2+C,y(\frac{\pi}{2})⋅ sin(\frac{\pi}{2})= \frac{(\frac{\pi}{2})^2}{2} +C \Rightarrow \\1⋅ sin(\frac{\pi}{2})= \frac{1}{2} (\frac{\pi}{2})^2+C,


or

1=Ο€28+Cβ‡’C=1βˆ’Ο€28.1= \frac{\pi^2}{8}+C \Rightarrow C= 1- \frac{\pi^2}{8}.


Then we obtain the solution for the initial value problem

𝑦⋅sinπ‘₯=π‘₯22+1βˆ’Ο€28𝑦 β‹… sinπ‘₯ = \frac{π‘₯^2}{2} +1- \frac{\pi^2}{8}

or


𝑦=(π‘₯22+1βˆ’Ο€28)csc(x)𝑦 =( \frac{π‘₯^2}{2} +1- \frac{\pi^2}{8})csc(x)


The solution for the initial value problem is


𝑦=(π‘₯22+1βˆ’Ο€28)csc(x).𝑦 =( \frac{π‘₯^2}{2} +1- \frac{\pi^2}{8})csc(x).





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