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achacttn
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Hello. I am trying to solve this problem methodically but my solution does not seem to agree with the given answer.
The differential equation is:
(sinx)y' - (cosx)y = 1 + C
y' - (cosx/sinx)y = 1/sinx + C/sinx
When finding the integrating factor, I used:
e^-∫(cosx/sinx)dx
I was wondering if this was either
|sinx|^(-1) or -|sinx| and why? I tried solving the equation using both methods
1st method (using |sinx|^(-1))
(|sinx|^(-1))y' - cosx/((sinx)^2).y = 1/(sinx)^2 + C/(sinx)^2
LHS becomes
D[|sinx|^(-1).y] = 1/(sinx)^2 + C/(sinx)^2
Integrating both sides and multiplying by |sinx| gives
y = x/sinx + Cx/sinx + Dsinx
When using -|sinx| I get:
y = x/sinx Cx/sinx + D/sinx
The given solution is
y = -x.cosx + sinxln(|sinx|) - Acosx + Bsinx
I have no idea how to get this. Any help would be appreciated, thanks!
Edit: I see my error for using the |sinx|^(-1), I didn't integrate the right hand side properly.
I get |sinx|^(-1).y = -cotx + Ccotx + D
so
y = -xcosx + Ccosx + Dsinx, which is still not the right answer
Homework Statement
The differential equation is:
(sinx)y' - (cosx)y = 1 + C
Homework Equations
The Attempt at a Solution
y' - (cosx/sinx)y = 1/sinx + C/sinx
When finding the integrating factor, I used:
e^-∫(cosx/sinx)dx
I was wondering if this was either
|sinx|^(-1) or -|sinx| and why? I tried solving the equation using both methods
1st method (using |sinx|^(-1))
(|sinx|^(-1))y' - cosx/((sinx)^2).y = 1/(sinx)^2 + C/(sinx)^2
LHS becomes
D[|sinx|^(-1).y] = 1/(sinx)^2 + C/(sinx)^2
Integrating both sides and multiplying by |sinx| gives
y = x/sinx + Cx/sinx + Dsinx
When using -|sinx| I get:
y = x/sinx Cx/sinx + D/sinx
The given solution is
y = -x.cosx + sinxln(|sinx|) - Acosx + Bsinx
I have no idea how to get this. Any help would be appreciated, thanks!
Edit: I see my error for using the |sinx|^(-1), I didn't integrate the right hand side properly.
I get |sinx|^(-1).y = -cotx + Ccotx + D
so
y = -xcosx + Ccosx + Dsinx, which is still not the right answer
Last edited: