Arbitrary Constant

Question

using elimination of arbitrary constant, determine differential equations satisfied by the following functions: y=c1e^t sin 3t + c2 e^t cos 3t

a. y" + 2y' - 10y = 0

b. y" - 10y' + 2y = 0

c. 2y" - y' + 10y = 0

d. y" - 2y' + 10y = 0

Answer 1

A another way is using the determinant (D = a^2 - 4b) of the characteristic polynomial, r^2 + ar + b.

a) D = 4+40 > 0 so solution is not of sinusoidal form.

b) D = 100 - 4 > 0 so solution is not of sinusoidal form.

c) D = 1/4 - 20 < 0 so solution is of sinusoidal form.

d) D = 4 - 100 < 0 so solution is of sinusoidal form.

So, a) and b) cannot be satisfied by the function.

Now we use elimination of arbitrary constants.

y = e^t (c1 sin(3 t) + c2 cos(3 t))

y' = e^t ((c1 - 3 c2) sin(3 t) + (3 c1 + c2) cos(3 t))

y'' = 2 e^t ((3 c1 - 4 c2) cos(3 t) - (4 c1 + 3 c2) sin(3 t))

We can recollect in terms of c1 and c2, but I think the following will be easier.

By Euler's formula,

y = c1 e^((1 - 3 i) t) + c2 e^((1 + 3 i) t).

y' = (1 - 3 i) c1 e^((1 - 3 i) t) + (1 + 3 i) c2 e^((1 + 3 i) t).

y'' = (-8 - 6 i) c1 e^((1 - 3 i) t) - (8 - 6 i) c2 e^((1 + 3 i) t).

Then subtracting multiples of equations to eliminate c1, we get:

y' - (1 - 3 i) y = 6 i c2 e^((1 + 3 i) t).

y'' - (-8 - 6 i) y = 12 i c2 e^((1 + 3 i) t)

Now we subtract multiples of these equations to eliminate c2:

y'' - (-8 - 6 i) y - 2 (y' - (1 - 3 i) y) = 0

Simplifying:

y'' - 2 y' + 10 y = 0

This is choice d).