2nd order differential - particular solution

[W3bbo]

[SOLVED] 2nd order differential - particular solution

Homework Statement

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

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

Homework Equations

N/A

The Attempt at a Solution

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

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

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


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

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:

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

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!

 

[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.

 

[HallsofIvy]

Homework Statement

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

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

Homework Equations

N/A

The Attempt at a Solution

$$\[<br /> \begin{array}{c}<br /> 2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9 \\ <br /> \\ <br /> 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 [itex]d^2x/dt^2= m^2 e^{mt}<br /> [/quote]$$\begin{array} 2t^2 e^{mt} + 5te^{mt} + 2e^{mt} = 0 \\ <br /> e^{mt} \left( {2t^2 + 5t + 2} \right) = 0 \\ <br /> t_1 = - {\textstyle{1 \over 2}} \\ <br /> t_2 = - 2 \\ <br /> \end{array}<br /> \]$$<br /> t is your variable- you trying to find values for m, the coeficient in the exponent.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Complimentary:<br /> $$Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t}$$<br /> <br /> Particular:<br /> $$\begin{gathered}<br /> x = at + b;\frac{{dx}}<br /> {{dt}} = a;\frac{{d^2 x}}<br /> {{dt^2 }} = 0 \\ <br /> \\ <br /> 2\left( 0 \right) + 5\left( a \right) + 2\left( {at + b} \right) = 2t + 9 \\ <br /> 5a + 2at + 2b = \\ <br /> 2b + a\left( {5 + 2t} \right) = \\ <br /> 2at = 2t \\ \end{gathered}$$ </div> </div> </blockquote>[/itex]

[tex] I'm not sure why you would write it like that. 2at+ (5a+ 2b)= 2t+ 9 is simpler.<br /> <br /> $$\begin{gathered} a = 1 \\ <br /> \\ <br /> 5a + 2b = 9 \\ <br /> b = 2 \\ <br /> \therefore \\ <br /> x = Ae^{ - 2t} + Be^{ - \tfrac{1}<br /> {2}t} + t + 2 \\ <br /> \end{gathered} <br />$$<br /> <br /> <br /> That's my solution to Part A, Mathematica agrees with me when I run:<br /> <br /> <div class="bbCodeBlock bbCodeBlock--screenLimited bbCodeBlock--code"> <div class="bbCodeBlock-title"> <i class="fa--xf fal fa-code "><svg xmlns="http://www.w3.org/2000/svg" role="img" aria-hidden="true" ><use href="/data/local/icons/light.svg?v=1784150131#code"></use></svg></i> Code: </div> <div class="bbCodeBlock-content" dir="ltr"> <pre class="bbCodeCode" dir="ltr" data-xf-init="code-block" data-lang=""><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]</code></pre> </div> </div><br /> (Note y substituted for x, x substitued for t, C[1] for B, and C[2] for A.)[/tex]

[tex] Okay, so that is correct (since after finding "t₁" and "t₂" you actually used them for "m").<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> For part B, I got:<br /> <br /> $$\begin{gathered}<br /> x = Ae^{ - 2t} + Be^{ - \tfrac{1}<br /> {2}t} + t + 2 \\ <br /> 3 = Ae^{ - 2\left( 0 \right)} + Be^{ - \tfrac{1}<br /> {2}\left( 0 \right)} + 0 + 2 \\ <br /> 3 = A + B + 2 \\ <br /> 1 = A + B \\ <br /> \\ <br /> \frac{{dx}}<br /> {{dt}} = - 2Ae^{ - 2t} - \tfrac{1}<br /> {2}Be^{ - \tfrac{1}<br /> {2}t} + 1 \\ <br /> - 1 = - 2A - \tfrac{1}<br /> {2}B + 1 \\ <br /> \\ <br /> A = 1 - B \\ <br /> - 1 = - 2\left( {1 - B} \right) - \tfrac{1}<br /> {2}B + 1 \\ <br /> 0 = B \\ <br /> A = 1 \\ <br /> \\ <br /> \therefore \\ <br /> x = e^{ - 2t} + t + 2 \\ <br /> \end{gathered}$$ <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Yes, that is correct.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> However this doesn't work with the last question (C), since the minimum of that function isn't located at ~2.84<br /> <br /> Any ideas where I've gone wrong?<br /> <br /> Thanks! </div> </div> </blockquote> Unfortunately, you have left out the crucial calculations- why you say that doesn't work.<br /> 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 $x= 1/2+ ln(2)/2+ 2= 5/2+ ln(2)/2, exactly as given.$ </div> </div> </blockquote> </div> </div> </blockquote>[/tex]

 

[W3bbo]

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.