Differential equation- wordy question1- urgent

**kingkaisai2** · September 1st 2006, 06:53 PM

6. The number of people, x, in a queue at a travel centre t minutes after it opens is modelled by a differential equation dx/dt= 1.4t-0.5x for values of t up to 10. Sovlve the differential equation given that x=8 when t=o.

b. An alternative model gives the differential equation dx/dt= 1.4t-o.5x for the same values of t. Verify that x=13.6e^-0.5t+2.8t-5.6 satisfies this differential equation. Verify also that when t=0 this function takes the value 8.

Please give working and explantion

**CaptainBlack** · September 1st 2006, 11:08 PM

Originally Posted by kingkaisai2

6. The number of people, x, in a queue at a travel centre t minutes after it opens is modelled by a differential equation dx/dt= 1.4t-0.5x for values of t up to 10. Sovlve the differential equation given that x=8 when t=o.

You have:

$ \frac{dx}{dt}=1.4t-0.5x $

with initial condition $x=8$ when $t=0$ .

Now rewrite the DE as:

$ \frac{dx}{dt}+0.5x=1.4t\ \ \ \ \ \dots (1) $

and we see that we have an inhomogeneous linear ODE with constant
coefficients, and so a general solution is the sum of the general solution
of the homogeneous equation:

$ \frac{dx}{dt}+0.5x=0\ \ \ \ \ \dots (2) $ ,

and any particular integral of the original equation (1).

Now the homogeneous equation (2) can be solved by a number of methods,
we can use a trial solution $x(t)=A e^{\alpha t}$ , substitute this into
the equation and find the value of $\alpha$ that gives the correct answer.
Another method is to note that (2) is of variable separable type.

I will use the first of these methods, I will suppose that: $x(t)=A e^{\alpha t}$ , then:

$ \frac{dx}{dt}+0.5x=\alpha A e^{\alpha t} + 0.5\ A e^{\alpha t}=0 $ ,

so if this is a solution we must have $\alpha=-0.5$ , and the general
solution of (2) is: $x(t)=A e^{-0.5 t}$ .

Now we need to find a particular integral for (1). Here we note that the RHS is
a multiple of $t$ , so if $x=m t +c$ the LHS can be made to equal
the RHS with suitable choice of $m$ and $c.$ , so substituting this into (1) we get:

$ \frac{dx}{dt}+0.5x=m+0.5(mt+c)=1.4 t $ .

So equating the coefficient of $t$ in the above gives $0.5 m=1.4$ , so $m=2.8$ ,
and the constant terms gives $m+0.5 c =0$ , or $c=-5.6$ .

Hence a particular integral of (1) is:

$ x(t)=2.8 t-5.6 $ .

Combining this with the general solution of (2) found earlier we get the
general solution of (1) is:

$ x(t)=A e^{-0.5 t}+2.8 t-5.6\ \ \ \ \ \dots (3) $ .

Now we need to fit the initial conditions. Putting $t=0$ into (3)
we have:

$ x(0)=A -5.6=8 $

which we solve to find that $A=13.6$ is applicable for this problem.

RonL

**CaptainBlack** · September 1st 2006, 11:38 PM

Originally Posted by kingkaisai2

An alternative model gives the differential equation dx/dt= 1.4t-o.5x for the same values of t. Verify that x=13.6e^-0.5t+2.8t-5.6 satisfies this differential equation. Verify also that when t=0 this function takes the value 8.

Do the last part first, substitute $t=0$ into the equation:

$ x(t)=13.6 e^{-0.5 t}+2.8 t-5.6\ \ \ \ \ \dots(A) $

to get:

$ x(0)=13.6 e^{-0.5\times 0}+2.8\times 0-5.6=13.6-5.6=8 $ ,

as required.

Now to verify that (A) satisfies the DE we differentiate $x(t)$ , and check
that this derivative equals $1.4 t-0.5x$ , which I leave as an exercise for the reader.

RonL

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