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:

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

$(du)/dx = -2 cotu$