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# Calculus Integration --> Long answer*

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

It takes half a page to solve, I heard.

The simplest way is by direct substitution. You are given a solution. Just plug it into the given differential equation, evaluate, group terms with y, y', and y''; and simplify.  It is a tedious process.

### 1 Answer by Expert Tutors

Andy C. | Math/Physics TutorMath/Physics Tutor
4.9 4.9 (21 lesson ratings) (21)
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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.