Differential Equations: Solve the following

[komarxian]

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

[/B]
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?

 

[Ray Vickson]

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

[/B]
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
View attachment 228687

any recommendations as to where I went wrong?

If $y'(x) + p(x) y(x) = q(x),$ the solution is of the form
$$y(x) = e^{-P(x)} \int_0^x q(t) e^{P(t)} \, dt + c e^{-P(x)},$$
where $P(w) = \int_0^w p(u) \, du$ for any $w$.

Note that if we include another constant of integration alongside $P$ (so we replace $P(w)$ by $P(w)+k$ for a constant $k$) the constant cancels in the first term, and replaces $c$ by $c e^{-k} = \text{another constant}$ in the second term.

 

[vela]

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)

You made a sign error when finding the integrating factor $p(x)$. If you differentiate $\log \cos x$, you get $-\tan x$.

Also, if you differentiate $y \cos x$, you get $(\cos x) y' + (-\sin x) y$, which isn't the lefthand side of the DE you started with.