Question

Solve the following differential equations using classical methods. Assume zero initial conditions.

Answer 1

I'll use the integrating factor method for the first DE, and undetermined coefficients for the second one.

(a) Multiply both sides by exp(7t ):

exp(7t ) dx/dt + 7 exp(7t ) x = 5 exp(7t ) cos(2t )

The left side is now the derivative of a product:

d/dt [exp(7t ) x] = 5 exp(7t ) cos(2t )

Integrate both sides:

exp(7t ) x = 10/53 exp(7t ) sin(2t ) + 35/53 exp(7t ) cos(2t ) + C

Solve for x :

x = 10/53 sin(2t ) + 35/53 cos(2t ) + C exp(-7t )

(b) Solve the corresonding homogeneous DE:

d²x/dt ² + 6 dx/dt + 8x = 0

has characteristic equation

r ² + 6r + 8 = (r + 4) (r + 2) = 0

with roots at r = -4 and r = -2. So the characteristic solution is

x (char.) = C₁ exp(-4t ) + C₂ exp(-2t )

For the particular solution, assume an ansatz of the form

x (part.) = a cos(3t ) + b sin(3t )

with derivatives

dx/dt = -3a sin(3t ) + 3b cos(3t )

d²x/dt ² = -9a cos(3t ) - 9b sin(3t )

Substitute these into the non-homogeneous DE and solve for the coefficients:

(-9a cos(3t ) - 9b sin(3t ))

… + 6 (-3a sin(3t ) + 3b cos(3t ))

… + 8 (a cos(3t ) + b sin(3t ))

= (-a + 18b) cos(3t ) + (-18a - b) sin(3t ) = 5 sin(3t )

So we have

-a + 18b = 0

-18a - b = 5

==>   a = -18/65 and b = -1/65

so that the particular solution is

x (part.) = -18/65 cos(3t ) - 1/65 sin(3t )

and thus the general solution is

x (gen.) = x (char.) + x (part.)

x = C₁ exp(-4t ) + C₂ exp(-2t ) - 18/65 cos(3t ) - 1/65 sin(3t )