Question

solve the given initial - value problem.
\\(\frac{dy}{dx}+(\cos(x))y = 0\\), \\(y(\frac{7\pi}{6})=-2\\)
give the largest interval i on which the solution is defined. (enter your answer using interval notation.)

Explanation:

Step1: Identify the type of differential equation

The given differential equation $\frac{dy}{dx}+\sin(x)y = 0$ is a first - order linear ordinary differential equation of the form $\frac{dy}{dx}+P(x)y = 0$, where $P(x)=\sin(x)$.

Step2: Solve the differential equation

We can rewrite it as $\frac{dy}{y}=-\sin(x)dx$. Integrating both sides:
$$\int\frac{dy}{y}=-\int\sin(x)dx$$
The integral of $\frac{1}{y}$ with respect to $y$ is $\ln|y|$, and the integral of $\sin(x)$ with respect to $x$ is $-\cos(x)+C$. So, $\ln|y|=\cos(x)+C$.

Step3: Use the initial condition

Given $y(\frac{7\pi}{6})=- 2$. Substitute $x = \frac{7\pi}{6}$ and $y=-2$ into $\ln|y|=\cos(x)+C$.
$\ln(2)=\cos(\frac{7\pi}{6})+C$. Since $\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}$, then $C=\ln(2)+\frac{\sqrt{3}}{2}$.
So, $\ln|y|=\cos(x)+\ln(2)+\frac{\sqrt{3}}{2}$, and $y = 2e^{\cos(x)+\frac{\sqrt{3}}{2}}$.

Step4: Determine the domain

The function $y = 2e^{\cos(x)+\frac{\sqrt{3}}{2}}$ is defined for all real - valued $x$ since the exponential function $e^u$ is defined for all real $u$ and $\cos(x)$ is defined for all real $x$.

Answer:

$(-\infty,\infty)$