Question

bc unit 7 (ab topics) quiz ver a
(a) $\frac{5}{21}$
(b) $sqrt{33}$
(c) $4 + e^{4}$
(d) $5e^{4}$

1. if $\frac{dy}{dx}=16\sin^{3}x\cos x$ and $y = 16$ when $x=\frac{\pi}{2}$, what is the value of $y$ when $x=\frac{\pi}{6}$?

(a) $\frac{49}{4}$
(b) $\frac{1}{4}$
(c) $10$
(d) $13$

1. if $dy/dx = 2y^{2}$ and if $y=-1$ when $x = 1$, then when $x = 2$, $y=$

(a) $-2/3$
(b) $-1/3$
(c) $0$
(d) $1/3$
(e) $2/3$

1. if $\frac{dy}{dx}=4y$ and if $y = 4$ when $x = 0$, then $y=$

(a) $4e^{4x}$
(b) $e^{4x}$
(c) $3 + e^{4x}$
(d) $4 + e^{4x}$
(e) $2x^{2}+4$

Explanation:

Step - by - Step Format:

Step 1: Solve the differential equation \(\frac{dy}{dx}=4y\)

Separate the variables: \(\frac{dy}{y} = 4dx\)
Integrate both sides:
\(\int\frac{dy}{y}=\int4dx\)
Using the integral formulas \(\int\frac{1}{y}dy=\ln|y|+C_1\) and \(\int4dx = 4x + C_2\), we get \(\ln|y|=4x + C\) (where \(C = C_2 - C_1\))
Exponentiate both sides: \(y = e^{4x + C}=e^{C}e^{4x}\)

Step 2: Use the initial condition \(y = 4\) when \(x = 0\)

Substitute \(x = 0\) and \(y = 4\) into \(y=e^{C}e^{4x}\)
\(4=e^{C}e^{0}\)
Since \(e^{0}=1\), then \(e^{C}=4\)

Step 3: Write the particular solution

The particular solution is \(y = 4e^{4x}\)

Answer:

A. \(4e^{4x}\)