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anonymous

  • 5 years ago

suppose that customer demand depends upon the price trend according to the formula q=60-20p-7p'(t)+p"(t). if the supply function is qs (p)=-12+10p, write down the condition for equilibrium and determine equilibrium price p(t) when p(0)=5 and p'(0)=17

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  1. anonymous
    • 5 years ago
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    I'm not trained in economics, just maths/physics. I assume the condition for equilibrium is supply = demand, so\[-12+10p=60-20p-7p'+p''\]is what you have to solve. Clean this thing up to get,\[p''-7p'-30p+72=0\]To make this homogeneous, set\[p=-\frac{v-72}{30}\rightarrow p'=-\frac{v'}{30} \rightarrow p''=-\frac{v''}{30}\]and sub into the differential equation to eliminate p. After doing this, mutiply through by -30 to give,\[v''-7v'-30v=0\]This is second order, ordinary differential, which can be solved assuming solutions for the form\[v=e^{\lambda t}\]Substituting this assumed solution into the d.e. yields the characteristic function,\[\lambda^2-7\lambda-30\lambda=0\rightarrow \lambda =10, -3\]

  2. anonymous
    • 5 years ago
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    So your solution for v is\[v=c_1e^{10t}+c_2e^{-3t}\]But from our transformation, \[v=72p-30\]so\[72-30p=c_1e^{10t}+c_2e^{-3t} \rightarrow p=c_1e^{10t}+c_2e^{-3t}+\frac{72}{30}\]

  3. anonymous
    • 5 years ago
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    (note the constants have absorbed any other constants we've come across in solving for p). You should be able to do the rest. If not, let me know.

  4. anonymous
    • 5 years ago
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    To find the constants,\[p(0)=5=c_1+c_2+\frac{72}{30}\rightarrow c_1+c_2=-\frac{13}{5}\]

  5. anonymous
    • 5 years ago
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    \[p'(0)=17=10c_1-3c_2\]From the first condition,\[c_2=-\frac{13}{5}-c_1\]and subbing into the second condition,\[17=10c_1-3(-\frac{13}{5}-c_1)\rightarrow c_1=\frac{46}{65}\]and so\[c_2=-\frac{13}{5}-\frac{46}{65}=-\frac{43}{13}\]

  6. anonymous
    • 5 years ago
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    \[p=\frac{46}{65}e^{10t}-\frac{43}{13}e^{-3t}+\frac{72}{30}\]

  7. anonymous
    • 5 years ago
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    Hi I tried working it out last night my answer was similar to yours just a few differences in the method of working it out but thank you so much for your insight. I am now starting differential equations and i'm having a hard time catching on...:(

  8. anonymous
    • 5 years ago
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    No probs. It's just experience in recognition with these things (and some insight). If you could 'fan' me as thanks, I'd appreciate it - I need points :)

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