How Do You Solve This Trigonometric Differential Equation from Calculus II?

[sourcandy]

Homework Statement

y=sinxcosx-cos^{2}x Differential equation= 2y+y'=2sin(2x)-1
Initial condition y(pi/4)=0

The Attempt at a Solution

2y=2sinxcosx-2cos^{2}x

y'=-sin^{2}x+cos^{2}x+2cosxsinx

ok... using double angle formulas... I get
2y=Sin2x-2cos^{2}x

y'=2cos^{2}x-1+2sinx

2y+y'
(Sin2x-2Cos^{2}x)+(2cos^{2}x-1+2sinx)
the 2cos mess cancells out leaves you with

Sin2x + (-1) +2Sinx
I don't understand how that equals 2sin2x-1
as the book says it does and the solution shows.

Stupid question I know... I struggle with the easy stuff.

Thanks
-Ed

 

[konthelion]

You're trying to prove that $$sin(2x) + 2sin(x)-1 = 2sin(2x)-1$$

Edit: I have to leave in a hurry, but this equality is not true.
In the RHS, $$2sin(2x)-1=sin(2x)+sin(2x)-1$$

$$\implies sin(2x) + 2sin(x)-1 = sin(2x)+sin(2x)-1$$
Therefore, cancelling the sin(2x)-1's from the LHS and RHS,
$$2sin(x) \neq sin(2x)$$ since by the double angle formula,

$$sin(2x) = 2sin(x)cos(x)$$

So, you probably made a mistake somewhere before.

 

[sourcandy]

awesome... I'm totally screwed for tomorrow. lol

Cal 2 during the summer was a terrible idea.