Science:Math Exam Resources/Courses/MATH101/April 2010/Question 07

MATH101 April 2010

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Question 07
Find the solution of the differential equation ${\frac {dL}{dt}}=kL^{2}\ln t$ that satisfies $L(1)=1$. Here, $k$ is a constant that will appear in your final answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can you separate the variables $L$ and $t$?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Beginning with the equation ${\frac {dL}{dt}}=kL^{2}\ln t$ we separate the variables $L$ and $t$ to obtain ${\frac {1}{L^{2}}}dL=k\ln t\,dt$ If we then integrate both sides of this equation (ignoring the constant of integration in the first case) we find that $\int {\frac {1}{L^{2}}}dL=-{\frac {1}{L}}$ and using integration by parts we find that $\int k\ln t\,dt=k(t\ln t-t)+C$ and so if we set the two of these equal to each other we find that $-{\frac {1}{L}}=k(t\ln t-t)+C$ If we solve for $L$ we find $L=-{\frac {1}{k(t\ln t-t)+C}}$ Since $L(1)=1$ it follows that $C=k-1$ and so $L(t)=-{\frac {1}{k(t\ln t-t)+k-1}}$

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