Solve dy/dx + y*tan(x) = sec(x) given y(0) = 1.

Step 1: Identify the form of the differential equation

The given differential equation $\frac{dy}{dx}$ + y $\tan(x) = \sec(x)$ is a first-order linear differential equation. We can compare it to the standard form $\frac{dy}{dx}$ + P(x) $y = Q(x)$ to identify P(x) and Q(x).

Step 2: Calculate the integrating factor

To solve a first-order linear differential equation, we need to find the integrating factor (IF). The integrating factor is given by the formula $e^{\int P(x) dx}$. In this case, P(x) = $\tan(x)$, and the integral of $\tan(x)$ is $\log|\sec(x)|$. Therefore, the integrating factor simplifies to $\sec(x)$.

Step 3: Apply the solution formula

The general solution to a first-order linear differential equation is given by y $\cdot \text{IF} = \int Q(x) \cdot \text{IF}$ \, dx. We substitute Q(x) = $\sec(x)$ and the integrating factor $\sec(x)$ into this formula. The integral of $\sec^2(x)$ is $\tan(x)$, so we get y $\sec(x) = \tan(x) + C$ where C is the constant of integration.

Step 4: Solve for y

To find the explicit solution for y, we divide both sides of the equation by $\sec(x)$. We can rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}$ and $\sec(x)$ as $\frac{1}{\cos(x)}$ After simplification, we obtain y $= \sin(x) + C\cos(x)$.

Step 5: Apply the initial condition

We are given the initial condition y(0) = 1. We substitute $x = 0$ and $y = 1$ into the general solution $y = \sin(x) + C\cos(x)$. Since $\sin(0) = 0$ and $\cos(0) = 1$, we find that 1 = 0 + C(1), which means $C = 1$.

Step 6: Write the particular solution

Now that we have found the value of the constant $C = 1$, we substitute it back into the general solution to get the particular solution for the given initial condition. The particular solution is $y = \sin(x) + \cos(x)$.

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