- #1

tylersmith7690

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## Homework Statement

For the following differential equation:

dy/dx = [itex]\frac{2cos^2x-sin^2x+y^2}{2 cosx}[/itex] , -pi/2 < x < pi/2

show that the substitution y(x)=sin x + 1/u(x) yeilds the differential equation for u(x),

du/dx = -u tan x - [itex]\frac{1}{2}[/itex]sec x

Hence find the solution y(x) to the original differential equation that satisfies the condition y(0)=2

Find the interval on which the solution to the initial value problem is defined.

## Homework Equations

I have no idea where to go next to get y(x).

## The Attempt at a Solution

using y= sinx + 1/u

dy/dx = cos x - 1/u^2 du/dx

then let this dy/dx = dy/dx in the original equation and solve for du.

cos x [itex]\frac{-1}{u^2}[/itex]du/dx = [itex]\frac{2cos^2x-sin^2x+y^2}{2cosx}[/itex]

cos x[itex]\frac{-1}{u^2}[/itex]du/dx= [itex]\frac{2 cos^2x+ (2/u) sinx + 1/u^2}{2cosx}[/itex]

[itex]\frac{-1}{u^2}[/itex]du/dx = [itex]\frac{1}{u}[/itex][itex]\frac{sinx}{cosx}[/itex]+[itex]\frac{1}{u^2}[/itex][itex]\frac{1}{2cosx}[/itex]

times through by -1/u^2

du/dx= -u tan x -1/2 sec x

Now du/dx + u tanx = -1/2 sec x, which is a first order linear equation

so integrating factor is I= sec x

so , sec x dx/du + u tanx sec x = -1/2 sec^2 x

so sec x u= [itex]\int -(1/2). sec^2 x[/itex]

= - 1/2 tan x +C

divide through by sec x

u= -1/2 sin x + C(cos x) [general solution]

now if i sub in u= 1/(y-sin x)

i cant rearrange it to get y by itself. This is where im stuck.

Sorry for the poor latex use. I have little knowledge of it atm.