stuck trying to integrate dy/dxy=cos(x)2by dooogle Tags: differential eq, integrating factor, integration by parts 

#1
Nov1110, 04:19 PM

P: 21

1. The problem statement, all variables and given/known data
integrate dy/dxy=cos(x)2 2. Relevant equations 3. The attempt at a solution dy/dxy=cos(x)2 is in the form dy/dx+p(x)=q(x) take the I.F as e^int(p(x))dx=e^x multiplying throughout by e^x d(e^x)z/dx=(cos(x)e^x)(2e^x) (e^x)z=int((cos(x)e^x)(2e^x))dx so int((cos(x)e^x)(2e^x))dx=int(cos(x)(e^x)dx2int(e^x)dx 2int(e^x)dx=2(e^x)+c int(cos(x)(e^x)) this is where i have trouble im not sure if my method is correct using integration by parts multiple times 1st integration let u=e^(x) du/dx=e^(x) dv/dx=cos(x) v=sin(x) int(cos(x)(e^x))dx = e^(x)sin(x)+int(sin(x)e^(x))dx 2nd integration int(sin(x)e^(x))dx let u=e^(x) du/dx=e^(x) dv/dx=sin(x) v=cos(x) e^(x)cos(x)+int(cos(x)(e^x))dx so int(cos(x)(e^x)dx=e^(x)sin(x)+e^(x)cos(x)+int(cos(x)(e^x))dx this is where im not sure i take away from both sides int(cos(x)(e^x))dx giving 2int(cos(x)(e^x)dx=e^(x)sin(x)+e^(x)cos(x) int(cos(x)(e^x)dx=e^(x)/2(sin(x)+cos(x)) so (e^x)z=2(e^x)+c+e^(x)/2(sin(x)+cos(x))+c z=2+(sin(x)+cos(x))/2+c/(e^x) 1/y=2+(sin(x)+cos(x))/2+c/(e^x) y=1/(2+(sin(x)+cos(x))/2+c/(e^x)) thank you for your time to recap my problems is: im not sure if i am allowed to add integrals cheers dooogle 



#2
Nov1110, 07:49 PM

Sci Advisor
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P: 2,606

It's easier to do that integral by using the Euler formula
[tex]e^{\pm ix} = \cos x \pm i \sin x[/tex] to obtain an expression for [tex]\cos x[/tex] in terms of exponentials. However, your integration is correct if you mean int(cos(x)(e^x)dx= (e^(x)/2 ) (sin(x)+cos(x)) +c 



#3
Nov1110, 09:40 PM

Math
Emeritus
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Thanks
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P: 38,877

Sorry but the use of two "/"s in



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