Solving Diff EQ: y'=2cos^2x-sin^2x+y^2 w/y(0)=2

[tylersmith7690]

Homework Statement

For the following differential equation:

dy/dx = $\frac{2cos^2x-sin^2x+y^2}{2 cosx}$ , -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 - $\frac{1}{2}$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 $\frac{-1}{u^2}$du/dx = $\frac{2cos^2x-sin^2x+y^2}{2cosx}$

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

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

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= $\int -(1/2). sec^2 x$

= - 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 can't rearrange it to get y by itself. This is where I am stuck.

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

 

[Simon Bridge]

You have a solution: $u(x)=\frac{1}{2}\sin x + C\cos x$
You want to sub in: $u(x)=1/(y(x)-\sin x)$
... and make y(x) the subject?

I don't see the problem - blind substitution gives you:
$$\frac{1}{y(x)-\sin x} = \frac{1}{2}\sin x + C\cos x$$ ... you got this far right?
Put the RHS under a common denominator then invert both sides.
Note: this is also a LaTeX lesson ;)

 

[tylersmith7690]

I don't see how you get y by itself. Maybe my algebra is way off. And i believe the RHS should have a negative before the 1/2 sin x.

Any help would be welcomed.

 

[Simon Bridge]

Please note: I am not allowed to do the working for you.
Put the RHS under a common denominator then invert both sides - as suggested in post #2.
If you are still stuck, please write down the working you have done in your reply.

Exercise:
Make y the subject in this example:
$$\frac{1}{y+5}=\frac{x}{2}+2$$

 

[tylersmith7690]

1/ (y-sinx) = -sinx/2 + C cosx

= (-sinx +2 C cos x)/2

y-sinx = 2/(-sinx +2 C cos x)

y= 2/(-sinx +2 C cos x) +sin x

y= [2 + sinx ( 2 cosx - sin x)]/ 2 cos x - sin x

y = (2 + 2 C sin x cos x -sin^2 x ) / ( 2 cos x - sin x)

y= 2+ C sin 2x - sin^2x / 2 cos x - sinx

This obviously can't be true because i need to find c and can't do this if i sub y=2 and x=0. But yeah that's my attempt.

 

[Simon Bridge]

1/ (y-sinx) = -sinx/2 + C cosx

= (-sinx +2 C cos x)/2

y-sinx = 2/(-sinx +2 C cos x)

y= 2/(-sinx +2 C cos x) +sin x

Whew you made it!
Why did you keep going?

y= [2 + sinx ( 2 cosx - sin x)]/ 2 cos x - sin x

y = (2 + 2 C sin x cos x -sin^2 x ) / ( 2 cos x - sin x)

y= 2+ C sin 2x - sin^2x / 2 cos x - sinx

This obviously can't be true because i need to find c and can't do this if i sub y=2 and x=0.

That's because you left out a C in the denominator in the last two lines. You left it out completely in the first line, and you forgot the brackets in the last line. When you are having trouble is the time to be extra careful about what you write down - take your time.

You don't need all that anyway...

Start with the step I put in boldface (above). Here it's formatted for you: $$y=\frac{2}{2C\cos x - \sin x}+\sin x$$ ... now find C given (x,y)=(0,2).
Hint: this is one of those situations where it is better to do the substitution before you do the algebra.