Where Did My Integration by Parts Go Wrong?

[Somefantastik]

x' = -2x + cos(t)

dx/dt + 2x = cos(t)

integrating factor = e^2t =>

e^{2t} \ \frac{dx}{dt} \ + 2 \ e^{2t} \ x \ = \ e^{2t} \cos(t) =>

\int \frac{d}{dt}\left(e^{2t} \ x \right) \ = \ \int e^{2t}\cos(t)\ dt =>

<br /> e^{2t}x = \int e^{2t}cos(t) \ dt = \frac{1}{2}cos(t) \ e^{2t} + \frac{1}{2}\int e^{2t}sin(t)\ dt = \frac{1}{2}cos(t)\ e^{2t} + \frac{1}{2}\left(-e^{2t}cos(t) + 2\int cos(t)\ e^{2t} \right) = \frac{1}{2}cos(t) \e^{2t} - \frac{1}{2}cos(t) \ e^{2t} + 2\int cos(t)\ e^{2t} dt \ =&gt;<br />

\int e^{2t} / cos(t) / dt = 0 + 2 \int cos(t) / e^{2t} / dt =&gt;

?? My integration must have gone wrong somewhere. Any explanations?

I was shown on an earlier thread how to transform the trig function into its complex form and solve that way, and the answer that resulted was that which agreed with good ole MATLAB : (1/5)*(2*cos(t) + sin(t)).

I'm grateful to be shown how to do this by using the transform, but my overwhelming curiosity pushes me to have this shown to me using integration by parts. I'm more likely to use that method on an exam. Any takers?

Thanks,
Candio

 

[HallsofIvy]

\int e^{2t}cos(t)dt
Let u= e^2t, dv= cos(t)dt.
Then du= 2e^2tdt, v= sin(t)
\int e^{2t}cos(t)dt= e^{2t}sin(t)- 2\int e^{2t}sin(t)dt
To do that second integral, let u= e^2t, dv= sin(t)dt.
Then du= 2e^2t, v= -cos(t)dt
\int e^{2t}sin(t)dt= -e^{2t}cos(t)+ 2\int e^{2t}cos(t)dt

Putting that into the previous integral,
\int e^{2t}cos(t)dt= e^{2t}sin(t)- 2(-e{2t}cos(t)+ 2\int e^{2t}cos(t)dt)
\int e^{2t}cos(t)dt= e^{2t}sin(t)+ 2e^{2t}cos(t)- 4\int e^{2t}cos(t)dt

Now add 4\int e^{2t}cos(t)dt to both sides of the equation.

 

[Somefantastik]

Ok thanks so much. When you have the type of int. by parts where you integrate by parts twice, must you be consistent when choosing u, dv?

 

[HallsofIvy]

Oh yes! If you switch in the middle, you will just reverse what you did the first time and everything will cancel!

 

[Somefantastik]

oh my goodness :(

thank you so much.