# 2nd order differential - particular solution

1. Jan 4, 2008

### W3bbo

[SOLVED] 2nd order differential - particular solution

1. The problem statement, all variables and given/known data

a) Find the general solution of the differential equation:

$$$2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9$$$

b) Find the particular solution of this differential equation for which:

$$$x = 3{\rm{ and }}\frac{{dy}}{{dx}} = - 1{\rm{ when }}t = 0$$$

c) The particular solution in part b) is used to model the motion of a particle P on the x-axis. At time t seconds (t>=0), P is x metres from the origin O. Show that the minimum distance between O and P is

$$${\textstyle{1 \over 2}}\left( {5 + {\mathop{\rm Ln}\nolimits} \left( 2 \right)} \right)$$$

2. Relevant equations

N/A

3. The attempt at a solution

$$$\begin{array}{c} 2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9 \\ \\ x = e^{mt} ;\frac{{dx}}{{dt}} = te^{mt} ;\frac{{d^2 x}}{{dt^2 }} = t^2 e^{mt} \\ \\ 2t^2 e^{mt} + 5te^{mt} + 2e^{mt} = 0 \\ e^{mt} \left( {2t^2 + 5t + 2} \right) = 0 \\ t_1 = - {\textstyle{1 \over 2}} \\ t_2 = - 2 \\ \end{array}$$$

Complimentary:
$$Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t}$$

Particular:
$$\begin{gathered} x = at + b;\frac{{dx}} {{dt}} = a;\frac{{d^2 x}} {{dt^2 }} = 0 \\ \\ 2\left( 0 \right) + 5\left( a \right) + 2\left( {at + b} \right) = 2t + 9 \\ 5a + 2at + 2b = \\ 2b + a\left( {5 + 2t} \right) = \\ 2at = 2t \\ a = 1 \\ \\ 5a + 2b = 9 \\ b = 2 \\ \therefore \\ x = Ae^{ - 2t} + Be^{ - \tfrac{1} {2}t} + t + 2 \\ \end{gathered}$$

That's my solution to Part A, Mathematica agrees with me when I run:

Code (Text):
In[1]:= DSolve[ 2y''[x] + 5y'[x] + 2y[x] == 2x+9, y[x], x  ]
Out[1]= y[x] -> 2 + x + e^(-x/2) * C[1] + e^(-2x) * C[2]
(Note y substituted for x, x substitued for t, C[1] for B, and C[2] for A.)

For part B, I got:

$$\begin{gathered} x = Ae^{ - 2t} + Be^{ - \tfrac{1} {2}t} + t + 2 \\ 3 = Ae^{ - 2\left( 0 \right)} + Be^{ - \tfrac{1} {2}\left( 0 \right)} + 0 + 2 \\ 3 = A + B + 2 \\ 1 = A + B \\ \\ \frac{{dx}} {{dt}} = - 2Ae^{ - 2t} - \tfrac{1} {2}Be^{ - \tfrac{1} {2}t} + 1 \\ - 1 = - 2A - \tfrac{1} {2}B + 1 \\ \\ A = 1 - B \\ - 1 = - 2\left( {1 - B} \right) - \tfrac{1} {2}B + 1 \\ 0 = B \\ A = 1 \\ \\ \therefore \\ x = e^{ - 2t} + t + 2 \\ \end{gathered}$$

However this doesn't work with the last question (C), since the minimum of that function isn't located at ~2.84

Any ideas where I've gone wrong?

Thanks!

Last edited: Jan 4, 2008
2. Jan 4, 2008

### rock.freak667

Find dx/dt and equate to zero and find the value for t that makes x a minimum(I got$\frac{1}{2}ln2$).
sub that value into the Particular solution and it should work out.

Last edited: Jan 4, 2008
3. Jan 5, 2008

### HallsofIvy

Staff Emeritus
I assume you mean dx/dt, not dy/dx.

No, $dx/dt= m e^{mt}$ and [itex]d^2x/dt^2= m^2 e^{mt}
[/quote]$$\begin{array} 2t^2 e^{mt} + 5te^{mt} + 2e^{mt} = 0 \\ e^{mt} \left( {2t^2 + 5t + 2} \right) = 0 \\ t_1 = - {\textstyle{1 \over 2}} \\ t_2 = - 2 \\ \end{array} \]$$
t is your variable- you trying to find values for m, the coeficient in the exponent.

Okay, so that is correct (since after finding "t1" and "t2" you actually used them for "m").

4. Jan 5, 2008

### W3bbo

My apologies, the way the question uses x for the function and t for the arg threw me off and I confused the derivative with the original equations.

Thing is, the minimum point on the plot is below zero, but isn't a negative distance still a positive displacement? So the minimum points would be zero and around 0.75. So isn't the question a little wrong in this respect?

But yours (and now my) answers are in line with the answer given, so thanks.