Solve the initial-value problem for x as a function of t .

Question

Solve the initial-value problem for x as a function of t .

(t^2−10t+24)dx/dt=1,(t > 6, x(7)=0)

Answer 1

Hi Gabe!

Thanks for your question.

The first step to solving this differential equation is to recognize that it is a separable equation. This means we can collect all the x variables on one side of the equation and all the t variables on the other side of the equation, like so:

(t²−10t+24)dx/dt = 1,

dx/dt = 1/(t²-10t+24),

dx = dt/(t²-10t+24).

Now we compute the indefinite integral on both sides of the equation (careful to include an arbitrary constant of integration C):

∫dx = ∫dt/(t²-10t+24) + C

x = ∫dt/(t²-10t+24) + C.

To evaluate the indefinite integral on the RHS of this equation, we need to use partial fractions. This requires us to compute the partial fraction decomposition of 1/(t²-10t+24):

1/(t²-10t+24) = 1/((t-4)(t-6)),

= A/(t-4) + B/(t-6).

Multiplying both sides of the second equality above, we get

1 = A(t-6) + B(t-4).

Plugging in t = 4, we get 1 = -2A, so A = -1/2. Plugging in t = 6, we get 1 = 2B, so B = 1/2. Therefore, our partial fraction decomposition is

1/(t²-10t+24) = (-1/2)/(t-4) + (1/2)/(t-6),

= 1/2(1/(t-6) - 1/(t-4)).

Returning to our equation for x,

x = ∫dt/(t²-10t+24) + C,

= 1/2 ∫(1/(t-6) - 1/(t-4)) dt + C,

= 1/2 ln|t-6| - ln|t-4| + C,

= 1/2 ln|(t-6)/(t-4)| + C.

To determine the value of C, we use the initial condition x(7) = 0. Then,

x(7) = 1/2 ln|(7-6)/(7-4)| + C,

= 1/2ln(1/3) + C = 0 ---> C = -1/2ln(1/3), or C = 1/2ln(3), using properties of ln.

Substituting this value of C into our expression for x as a function of t, we get

x = 1/2 ln|(t-6)/(t-4)| + 1/2ln(3),

= 1/2 ln(3|(t-6)/(t-4)|),

where we've combined the ln's by the multiplication property of ln.