Question

shown above is a slope field for the differential equation \\( \frac{d y}{d x}=y^{2}\left(4-y^{2}\
ight) \\). if \\( y=g(x) \\) is the solution to the differential equation with the initial condition \\( g(-2)=-1 \\), then, \\( \lim _{x \
ightarrow \infty} g(x) \\) is
(a) \\( -\infty \\)
(b) -2
(c) 0
(d) 2
(e) 3

1. which of the following is the solution to the differential equation \\( \frac{d y}{d x}=y \cos x \\) with the initial condition \\( y\left(\frac{\pi}{2}\

ight)=e \\) ?
(a) \\( y=\cos x+e \\)
(b) \\( y=e^{\cos x+1} \\)
(c) \\( y=e \cdot \sin x \\)
(d) \\( y=e^{\sin x} \\)
\\( \frac{d y}{d x}=2 x y \\) with the initial condition \\( f(0)=5 \\). what is the value of

Explanation:

Step1: Solve the differential equation

We have the differential equation \(\frac{dy}{dx}=y\cos x\).
Separate the variables: \(\frac{dy}{y}=\cos xdx\).
Integrate both sides:
\(\int\frac{dy}{y}=\int\cos xdx\).
Using the integral formulas \(\int\frac{1}{y}dy=\ln|y| + C_1\) and \(\int\cos xdx=\sin x + C_2\), we get \(\ln|y|=\sin x + C\) (where \(C = C_2 - C_1\)).
Exponentiating both sides gives \(y = e^{\sin x + C}=e^{C}e^{\sin x}\).
Let \(k = e^{C}\), so \(y = ke^{\sin x}\).

Step2: Apply the initial condition

We are given the initial condition \(y(\frac{\pi}{2})=e\).
Substitute \(x = \frac{\pi}{2}\) and \(y = e\) into \(y = ke^{\sin x}\).
When \(x=\frac{\pi}{2}\), \(\sin x = 1\), so \(e=ke^{1}\).
Solving for \(k\), we divide both sides by \(e\): \(k = 1\).

Answer:

D. \(y = e^{\sin x}\)