Question

find the general solution of the given differential equation. $\frac{dy}{dx}+left(sin(x)
ight)y = 1$

Explanation:

Step1: Identify the type of differential equation

The given equation $\cos(x)\frac{dy}{dx}+\sin(x)y = 1$ is a first - order linear ordinary differential equation of the form $\frac{dy}{dx}+P(x)y = Q(x)$, where $P(x)=\tan(x)$ and $Q(x)=\sec(x)$ after dividing by $\cos(x)$ ($\cos(x)
eq0$).

Step2: Find the integrating factor

The integrating factor $\mu(x)=e^{\int P(x)dx}$. Here, $\int\tan(x)dx=\int\frac{\sin(x)}{\cos(x)}dx=-\ln|\cos(x)|=\ln|\sec(x)|$, so $\mu(x)=\sec(x)$.

Step3: Multiply the entire equation by the integrating factor

Multiply $\cos(x)\frac{dy}{dx}+\sin(x)y = 1$ by $\sec(x)$. We get $\sec(x)\cos(x)\frac{dy}{dx}+\sec(x)\sin(x)y=\sec(x)$, which simplifies to $\frac{dy}{dx}+\tan(x)y=\sec(x)$. The left - hand side is the derivative of the product $y\sec(x)$ by the product rule. So, $\frac{d}{dx}(y\sec(x))=\sec(x)$.

Step4: Integrate both sides

Integrate $\frac{d}{dx}(y\sec(x))=\sec(x)$ with respect to $x$. We know that $\int\sec(x)dx=\ln|\sec(x)+\tan(x)| + C$. So, $y\sec(x)=\ln|\sec(x)+\tan(x)|+C$.

Step5: Solve for y

Multiply both sides by $\cos(x)$ to get $y = \cos(x)\ln|\sec(x)+\tan(x)|+C\cos(x)$.

Answer:

$y = \cos(x)\ln|\sec(x)+\tan(x)|+C\cos(x)$