# Solving a Differential Equation

## 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!

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dextercioby
Homework Helper
Looks ok to me.

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

derivative y=√x+√x
y'=?

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...?

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
Gold Member
under sqrtx is also +sqrtx
Don't hijack other problems with your own.

Don't hijack other problems with your own.
sorry