Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 16 (b)

MATH101 A April 2024

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[hide] Question 16 (b)
Consider the separable differential equation $${\frac {\mathrm {d} y}{\mathrm {d} x}}-2xy+4y=8-4x,\quad -\infty <x<+\infty .$$ (b) A certain solution satisfies $y(0)=8.$ For this solution, what is the smallest value of $y(x)$ over all real $x$?

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!

[show] Hint
To find the specific solution, plug $x=0$ and $y=8$ into the general solution you found in part (a). Then find the smallest value $y$ that this specific solution takes.

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[show] Solution

In part (a), we found the general solution $${\begin{aligned}y=2+Ce^{x^{2}-4x}.\end{aligned}}$$ To find the value of $C$ such that $y(0)=8$, plug in $x=0$ and $y=8$: $${\begin{aligned}8=2+Ce^{0}=2+C.\end{aligned}}$$ The result is $C=6$. Therefore, the particular solution is $y=2+6e^{x^{2}-4x}$. Now we must find the smallest value $y$ that this solution takes. By completing the square, the exponent $x^{2}-4x$ is equal to $(x-2)^{2}-4$, which is minimized when $x=2$. Therefore, the smallest value $y$ takes is $y_{\mathrm {min} }=y(2)=2+6e^{-4}$.