Solving a Differential Equation


by gmmstr827
Tags: differential, equation, solving
gmmstr827
gmmstr827 is offline
#1
Feb6-11, 05:21 PM
P: 86
1. The problem statement, all variables and given/known data

Solve:
2 * √(x) * (dy/dx) = cos^2(y)
y(4) = π/4

2. Relevant equations

TRIGONOMETRIC RECIPROCAL IDENTITY: sec(u) = 1 / cos(u)
arctan( 1 ) = π / 4

3. The attempt at a solution

This is a separable differential equation.

2 * √(x) * (dy/dx) = cos^2(y)
2 * √(x) * dy = cos^2(y) * dx
[2 / cos^2(y)] * dy = [1 / √(x)] * dx
TRIGONOMETRIC RECIPROCAL IDENTITY: sec(u) = 1 / cos(u)
[2 * sec^2(y)] * dy = x^(-1/2) * dx

∫ [2 * sec^2(y)] * dy = ∫ x^(-1/2) * dx
2 * tan(y) = 2 * √(x) + C
y(x) = arctan( √(x) + C) <<< General Solution

NOTE: arctan( 1 ) = π / 4
y(4) = arctan( √(4) + C )
y(4) = arctan( 2 + C)
C = -1

y(x) = arctan( √(x) - 1 ) <<< Particular Solution

Is that all correct? Thank you!
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dextercioby
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#2
Feb6-11, 05:26 PM
Sci Advisor
HW Helper
P: 11,866
Looks ok to me.
gmmstr827
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#3
Feb6-11, 05:36 PM
P: 86
Thanks! I just remembered that I can check these in my calculator as well.... >.<

tinala
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#4
Feb6-11, 07:19 PM
P: 8

Solving a Differential Equation


derivative y=√x+√x
y'=?
gmmstr827
gmmstr827 is offline
#5
Feb6-11, 07:27 PM
P: 86
Quote Quote by tinala View Post
derivative y=√x+√x
y'=?
Um... what?

Well, if
y=√(x)+√(x)
then
y=2√(x)
and
y'=4x^(3/2)/3

but you need to separate the variables first, then integrate not derive, so I don't see how that's relevant...?
tinala
tinala is offline
#6
Feb7-11, 03:10 AM
P: 8
Quote Quote by gmmstr827 View Post
Um... what?

Well, if
y=√(x)+√(x)
then
y=2√(x)
and
y'=4x^(3/2)/3

but you need to separate the variables first, then integrate not derive, so I don't see how that's relevant...?
under sqrtx is also +sqrtx
Char. Limit
Char. Limit is offline
#7
Feb7-11, 03:11 AM
PF Gold
Char. Limit's Avatar
P: 1,930
Quote Quote by tinala View Post
under sqrtx is also +sqrtx
Don't hijack other problems with your own.
tinala
tinala is offline
#8
Feb7-11, 03:15 AM
P: 8
Quote Quote by Char. Limit View Post
Don't hijack other problems with your own.
sorry


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