[SOLVED] 2nd order differential - particular solution

W3bbo

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

a) Find the general solution of the differential equation:

[tex]\[2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9\]
[/tex]

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

[tex]\[x = 3{\rm{ and }}\frac{{dy}}{{dx}} = - 1{\rm{ when }}t = 0\][/tex]

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

[tex]\[{\textstyle{1 \over 2}}\left( {5 + {\mathop{\rm Ln}\nolimits} \left( 2 \right)} \right)\][/tex]

2. Relevant equations

N/A

3. The attempt at a solution

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

Complimentary:
[tex]Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t} [/tex]

Particular:
[tex]
$\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} $

[/tex]


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

(Note y substituted for x, x substitued for t, C[1] for B, and C[2] for A.)

For part B, I got:

[tex]
$\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} $
[/tex]

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!

rock.freak667

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

HallsofIvy

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

No, [itex]dx/dt= m e^{mt}[/itex] and [itex]d^2x/dt^2= m^2 e^{mt}
[/quote][tex]\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}
\][/tex]
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").

[quote]For part B, I got:

[tex]
$\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} $
[/tex][quote]
Yes, that is correct.

Unfortunately, you have left out the crucial calculations- why you say that doesn't work.
With [itex]x= e^{-2t}+ t+ 2[/itex]. [itex]dx/dt= -2e^{-2t}+ 1[/itex] and that equals 0 when [itex]e^{-2t)= 1/2[/itex] so t= ln(2)/2. Putting that into [itex]x= e^{-2t}+ t+ 2[/itex], we have [itex]x= 1/2+ ln(2)/2+ 2= 5/2+ ln(2)/2, exactly as given.

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.