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W3bbo
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[SOLVED] 2nd order differential - particular solution
a) Find the general solution of the differential equation:
\[2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9\]<br />
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
\[{\textstyle{1 \over 2}}\left( {5 + {\mathop{\rm Ln}\nolimits} \left( 2 \right)} \right)\]
N/A
<br /> \[<br /> \begin{array}{c}<br /> 2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9 \\ <br /> \\ <br /> x = e^{mt} ;\frac{{dx}}{{dt}} = te^{mt} ;\frac{{d^2 x}}{{dt^2 }} = t^2 e^{mt} \\ <br /> \\ <br /> 2t^2 e^{mt} + 5te^{mt} + 2e^{mt} = 0 \\ <br /> e^{mt} \left( {2t^2 + 5t + 2} \right) = 0 \\ <br /> t_1 = - {\textstyle{1 \over 2}} \\ <br /> t_2 = - 2 \\ <br /> \end{array}<br /> \]
Complimentary:
Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t}
Particular:
<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 \\ <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 /> <br />
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:
<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} $<br />
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!
Homework Statement
a) Find the general solution of the differential equation:
\[2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9\]<br />
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 /> \[<br /> \begin{array}{c}<br /> 2\frac{{d^2 x}}{{dt^2 }} + 5\frac{{dx}}{{dt}} + 2x = 2t + 9 \\ <br /> \\ <br /> x = e^{mt} ;\frac{{dx}}{{dt}} = te^{mt} ;\frac{{d^2 x}}{{dt^2 }} = t^2 e^{mt} \\ <br /> \\ <br /> 2t^2 e^{mt} + 5te^{mt} + 2e^{mt} = 0 \\ <br /> e^{mt} \left( {2t^2 + 5t + 2} \right) = 0 \\ <br /> t_1 = - {\textstyle{1 \over 2}} \\ <br /> t_2 = - 2 \\ <br /> \end{array}<br /> \]
Complimentary:
Ae^{ - 2t} + Be^{ - {\textstyle{1 \over 2}}t}
Particular:
<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 \\ <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 /> <br />
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:
<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} $<br />
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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