Science:Math Exam Resources/Courses/MATH103/April 2015/Question 01 (d) (i)

MATH103 April 2015

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Question 01 (d) (i)
Consider the following differential equations. Check whether the statement is true or false. (i) $x(t)=-{\frac {2}{t^{2}+2}}$ is a solution to ${\frac {dx}{dt}}=tx^{2}$.

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
Find ${\frac {dx}{dt}}$, substitute ${\frac {dx}{dt}}$ and $x$ into the equation and check whether the equation holds.

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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. If $x(t)=-{\frac {2}{t^{2}+2}}=-2(t^{2}+2)^{-1}$ then we use the chain rule to find ${\frac {dx}{dt}}$. ${\frac {dx}{dt}}=x'(t)=(-2)\cdot (-1)\cdot (t^{2}+2)^{-2}\cdot 2t=4t(t^{2}+2)^{-2}$ On the other hand, $tx^{2}=t\cdot \left(-2(t^{2}+2)^{-1}\right)^{2}=t\cdot 4(t^{2}+2)^{-2}=4t(t^{2}+2)^{-2}$, so we see that $\color {blue}{\frac {dx}{dt}}=tx^{2}$.