Calculus Integration --> Long answer*

Question

Picture of the question: http://imgur.com/a/J7QEq

 

It takes half a page to solve, I heard.

Answer 1

 

Here's a quick and dirty proof.

I skipped a lot of steps, especially on the chain rule.

For example, taking derivatives of cos, you get -sin.

Rather than lug the negative sign around I switched

the order of the coefficient terms (multiplying them by -1).

But it works.

 

After calculating the derivatives, I make a table

of just the coefficients of the sine and cosine terms.

It shall suffice to show those coefficients are zero

since the exponential term factors and cancels.

-------------------------------------------------------

y' = exp(-ax) * [ wl cos(wx) - kw sin(wx) ] + exp(-ax) [ -ak cos(wx) - al sin(wx) ]

= exp(-ax) * [ (wl - ak) cos(wx) - (kw + al) sin(wx) ]

--------------------------------------------------------------------

y'' = exp(-ax) * [ w(ak - wl) sin(wx) - w(kw + al) cos(wx)] +

exp(-ax) * [ a(ak -wl) cos(wx) + a(kw + al) sin(wx) ]


= exp(-ax) * [ (wak-w*l^2 + akw + a^2*l) sin(wx) + (a^2k - awl - kw^2 - wal) cos(wx)]

--------------------------------------------------------------------------------

Post plugin, the exponential, which is always positive, will factor out , cancel, and divide out.
So the trig functions MUST vanish.

Setting up a table of the coefficients for the sine and cosine terms:


sin(wx)                                              cos(wx)
-------------------------------------------------------------------

(wak-w*l^2 + akw + a^2*l)                    (a^2k - awl - kw^2 - wal) <---- y''


-(2a)(kw +al)                                           (2a)(wl - ak) <---- (2a)y'


(w^2+a^2)l                                                 (w^2 +a^2)k <--- (w^2 +a^2)y




For sin(wx):

wak-w*l^2 + akw + a^2*l - 2akw - 2a^2*l + w^2*l + a^2*l

a^2*l terms cancel. The remaining terms:
wak - w*l^2 + awk - 2awk + w^2*l

The awk terms cancel, as do the w*l^2 terms.
The coefficient of sin(wx) is zero.


For cos(wx):

a^2k - awl - kw^2 - wal + 2wal - 2a^2*k + k*w^2 + k*a^2

a^2*k terms cancel. THe remaining terms:
-awl - k*w^2 - wal + 2wal + k*w^2

The awl terms cancel, as do the k*w^2 terms.
The coefficient of cos(wx) is zero.


Done deal.