Answer to Question #96321 in Differential Equations for Olajide Olaitan

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


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


This equation is a linear equation of the form


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


where

"P(x) = cot (x),\ud835\udc44(\ud835\udc65) = \ud835\udc65 csc(x)"


Integrating factor :


"IF(x) = e^ {\\int \ud835\udc43(\ud835\udc65)\ud835\udc51x }=e^ {\\int cot (x) \ud835\udc51x }=\\\\ =e^ {\\int \\frac{\ud835\udc51(sin\ud835\udc65)\n}{sin x} }= e^ {ln|sin\ud835\udc65| + \ud835\udc36}"


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


Then we get 


"IF (x)= sin(\ud835\udc65)"


Multiply both sides of the equation by "IF (x)= sin(\ud835\udc65)"


"sin(\ud835\udc65) \\frac{dy}{dx} +(sin(\ud835\udc65) \u22c5cot(x))y=x csc(x)\u22c5 sin(\ud835\udc65)"

or


"sin(\ud835\udc65) \\frac{dy}{dx} + (cos(\ud835\udc65))\ud835\udc66 = \ud835\udc65"


i.e.


"\\frac{d}{dx}(\ud835\udc66 \u22c5 sin\ud835\udc65) = \ud835\udc65."


Integrate both sides of this equation we get the general solution


"\\int \\frac{d}{dx}(\ud835\udc66 \u22c5 sin\ud835\udc65)dx = \\int \ud835\udc65 dx \\Rightarrow \ud835\udc66 \u22c5 sin\ud835\udc65 = \\frac{\ud835\udc65^2}{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(\\frac{\\pi}{2}) =1"


into the general solution:


"y(\\frac{\\pi}{2})\u22c5 sin(\\frac{\\pi}{2})= \\frac{(\\frac{\\pi}{2})^2}{2} +C \\Rightarrow \\\\1\u22c5 sin(\\frac{\\pi}{2})= \\frac{1}{2} (\\frac{\\pi}{2})^2+C,"


or

"1= \\frac{\\pi^2}{8}+C \\Rightarrow C= 1- \\frac{\\pi^2}{8}."


Then we obtain the solution for the initial value problem

"\ud835\udc66 \u22c5 sin\ud835\udc65 = \\frac{\ud835\udc65^2}{2} +1- \\frac{\\pi^2}{8}"

or


"\ud835\udc66 =( \\frac{\ud835\udc65^2}{2} +1- \\frac{\\pi^2}{8})csc(x)"


The solution for the initial value problem is


"\ud835\udc66 =( \\frac{\ud835\udc65^2}{2} +1- \\frac{\\pi^2}{8})csc(x)."





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