Show that the general solution of the equation y\" a2y = 0. Where a is a constan

Question

Show that the general solution of the equation y" a2y = 0. Where a is a constant, can he written cither as y(x) c1eax + c2e-ax or as y(x) = C1, cosh(ax) + C2 sinh(ax). We begin with the polynomial equation for y" - a2y = 0 and have r2 - a2 = 0. Solving for r, we have r2 - a2 and r = plusminus a. Let r1 = a and r2 = -a. We know that the general solution for P(D)y = 0 are the functions er1x,er2x,,Therefore, after substituting for r1 and r2 we will have the solutions y1 = eax and y2 = e-ax. Then, the general solution is y(x) - c1eax + c2e-ax and we have completed the first part of the proof. In terms of y(x) = C1 cosh (ax) + C2 sinh(ax), I am confused of where to start.

Explanation / Answer

The way to show that these are solutions are simply to plug in the different solutions into the above diff eqn

y(x) = c1ex + c2e-x

y'(x) = c1ex - c2e-x

y''(x) = c12ex + c22e-x

Now we can plug in this info to see if this satisfies the diff eqn

y'' - 2y = 0

(c12ex + c22e-x) - 2(c1ex + c2e-x) =? 0

c12ex + c22e-x - 2c1ex - 2c2e-x=? 0

0=0

Yes

Next let us look at the other solution

y(x) = c1cosh(x) + c2sinh(x)

y'(x) = c1sinh(x) + c2cosh(x)

y''(x) = c12cosh(x) + c22sinh(x)

Now we can plug in this info to see if this satisfies the diff eqn

y'' - 2y = 0

(c12cosh(x) + c22sinh(x)) - 2(c1cosh(x) + c2sinh(x)) =? 0

(c12cosh(x) + c22sinh(x)) - 2c1cosh(x) - c22sinh(x) =? 0

0=0

Yes

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