## [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:
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!
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 Recognitions: Homework Help 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.

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 Quote by W3bbo 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: $$$x = 3{\rm{ and }}\frac{{dy}}{{dx}} = - 1{\rm{ when }}t = 0$$$
I assume you mean dx/dt, not dy/dx.

 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 $$$\begin{array}{c} 2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9 \\ \\ x = e^{mt} ;\frac{{dx}}{{dt}} = te^{mt} ;\frac{{d^2 x}}{{dt^2 }} = t^2 e^{mt} \end{array}$$ No, $dx/dt= m e^{mt}$ and $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.  Complimentary: $$Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t}$$ Particular: $$\begin{gathered} x = at + b;\frac{{dx}} {{dt}} = a;\frac{{d^2 x}} {{dt^2 }} = 0 \\ \\ 2\left( 0 \right) + 5\left( a \right) + 2\left( {at + b} \right) = 2t + 9 \\ 5a + 2at + 2b = \\ 2b + a\left( {5 + 2t} \right) = \\ 2at = 2t \\ \end{gathered}$ I'm not sure why you would write it like that. 2at+ (5a+ 2b)= 2t+ 9 is simpler. [tex]\begin{gathered} a = 1 \\ \\ 5a + 2b = 9 \\ b = 2 \\ \therefore \\ x = Ae^{ - 2t} + Be^{ - \tfrac{1} {2}t} + t + 2 \\ \end{gathered}$$ That's my solution to Part A, Mathematica agrees with me when I run: Code: 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.)
Okay, so that is correct (since after finding "t1" and "t2" you actually used them for "m").

[quote]For part B, I got:

$$\begin{gathered} x = Ae^{ - 2t} + Be^{ - \tfrac{1} {2}t} + t + 2 \\ 3 = Ae^{ - 2\left( 0 \right)} + Be^{ - \tfrac{1} {2}\left( 0 \right)} + 0 + 2 \\ 3 = A + B + 2 \\ 1 = A + B \\ \\ \frac{{dx}} {{dt}} = - 2Ae^{ - 2t} - \tfrac{1} {2}Be^{ - \tfrac{1} {2}t} + 1 \\ - 1 = - 2A - \tfrac{1} {2}B + 1 \\ \\ A = 1 - B \\ - 1 = - 2\left( {1 - B} \right) - \tfrac{1} {2}B + 1 \\ 0 = B \\ A = 1 \\ \\ \therefore \\ x = e^{ - 2t} + t + 2 \\ \end{gathered}$$[quote]
Yes, that is correct.

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

## [SOLVED] 2nd order differential - particular solution

 Quote by HallsofIvy Unfortunately, you have left out the crucial calculations- why you say that doesn't work.
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.