answer: $\\frac{3}{2}$ # $y=\\sec^{2}x + 3x$, $y(0)=3$ to advance in the circuit, locate $y(\\frac{\\pi}{4})$.
answer: $\\frac{3}{2}$ # $y=\\sec^{2}x + 3x$, $y(0)=3$ to advance in the circuit, locate $y(\\frac{\\pi}{4})$.
Answer
Explanation:
Step1: Integrate (y')
We know that if (y' = f(x)), then (y=\int f(x)dx + C). Since (y'=\sec^{2}x + 3x), and (\int\sec^{2}x dx=\tan x), (\int3x dx=\frac{3x^{2}}{2}). So (y=\tan x+\frac{3x^{2}}{2}+C).
Step2: Find the constant (C)
Use the initial condition (y(0) = 3). Substitute (x = 0) and (y=3) into (y=\tan x+\frac{3x^{2}}{2}+C). We get (3=\tan(0)+\frac{3\times0^{2}}{2}+C), since (\tan(0) = 0), then (C = 3). So (y=\tan x+\frac{3x^{2}}{2}+3).
Step3: Calculate (y(\frac{\pi}{4}))
Substitute (x=\frac{\pi}{4}) into (y=\tan x+\frac{3x^{2}}{2}+3). We know that (\tan(\frac{\pi}{4})=1), (x = \frac{\pi}{4}), then (y=\tan(\frac{\pi}{4})+\frac{3}{2}\times(\frac{\pi}{4})^{2}+3). (y=1 + 3+\frac{3\pi^{2}}{32}=4+\frac{3\pi^{2}}{32})
Answer:
(4+\frac{3\pi^{2}}{32})