Science:Math Exam Resources/Courses/MATH101 B/April 2024/Question 16 (b)/Solution 1

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 $y_{\mathrm {min} }=y(2)=2+6e^{-4}$.