X (du/dx)=-2cot (u) , y=ux Find the solution of the differential equation satisfying the condition y=0 when x=1. Write the answer in the form y=f (x)?

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Answer 1 · 2018-04-17T12:43:09

#bary = x arccos(e^(2(x-1)))#

The differential equation:

#(du)/dx = -2cotu#

is separable:

#(du)/cot u = -2dx#

#int (sinu)/cosu du = -2int dx#

#- int (d(cosu))/cosu = -2int dx#

#ln abs cos u = 2x +C#

#cosu = ce^(2x)#

#u = arccos (ce^(2x))#

#y = ux = xarccos (ce^(2x))#

is the general solution.

For #x=1# we must have #y=0# and we can determine #c# from the resulting equation:

#0 = arccos(c e^2)#

so:

#c e^2 = 1#

#c= 1/e^2#

and then:

#bary = xarccos (e^(2x)/e^2) = x arccos(e^(2(x-1)))#

In fact:

#u = y/x = arccos(e^(2(x-1)))#

then:

#cos u = e^(2(x-1))#

#ln cosu = 2(x-1)#

and differentiating implicitly:

#-sinu/cosu (du) /dx = 2#

#(du)/dx = -2 cotu#