Solving a Differential Equation

[gmmstr827]

Homework Statement

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

Homework Equations

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

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!

 

[dextercioby]

Looks ok to me.

 

[gmmstr827]

Thanks! I just remembered that I can check these in my calculator as well... >.<

 

[tinala]

derivative y=√x+√x
y'=?

 

[gmmstr827]

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]

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]

under sqrtx is also +sqrtx

Don't hijack other problems with your own.

 

[tinala]

Don't hijack other problems with your own.

sorry