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## Homework Statement

Solve the following differential equations/initial value problems:

(cosx) y' + (sinx) y = sin2x

## Homework Equations

I've been attempting to use the trig ID sin2x = 2sinxcosx.

I am also trying to solve this problem by using p(x)/P(x) and Q(x)

## The Attempt at a Solution

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cosx y' +sinx y = 2sinxcosx

y' + tanx y = 2 sinx

P(x) = tanx, Q(x) = 2sinx

--> p(x) = e^lncosx + c = e^ lncosx e^c = C cos x (if e^c = C)

Dx = p(x)Q(x) = Ccosx (2sinx) = c 2 sinxcosx = Csin2x

p(x) y(x) = int( Dx ) dx = int( Csin2x) dx = C -cos2x/2 + C

y(x) = C -cos2x/2(Ccosx) + C/Ccosx = -Ccos2x/2(Ccosx) + 1/cosx = -Ccos2x/2(Ccosx) + secx

I'm kind of confused, because my constant C disappears at some point, and the answer is supposed to be

any recommendations as to where I went wrong?