Question

solve the given differential equation by using an appropriate substitution. the de is a bernoulli equation.
\\(\frac{dy}{dx}+y = \frac{1}{y}\\)

Explanation:

Step1: Rewrite the Bernoulli equation

The given differential equation $\frac{dy}{dx}+y = \frac{1}{x}$ is a first - order linear differential equation (it is in the form $\frac{dy}{dx}+P(x)y = Q(x)$ where $P(x) = 1$ and $Q(x)=\frac{1}{x}$). The integrating factor $\mu(x)$ is given by $e^{\int P(x)dx}$.

Step2: Calculate the integrating factor

We have $\int P(x)dx=\int 1dx=x$, so $\mu(x)=e^{x}$.

Step3: Multiply the entire equation by the integrating factor

Multiply $\frac{dy}{dx}+y=\frac{1}{x}$ by $e^{x}$ to get $e^{x}\frac{dy}{dx}+e^{x}y=\frac{e^{x}}{x}$. The left - hand side is the derivative of the product $e^{x}y$ by the product rule $(uv)^\prime = u^\prime v+uv^\prime$ (where $u = e^{x}$ and $v = y$), so $\frac{d}{dx}(e^{x}y)=\frac{e^{x}}{x}$.

Step4: Integrate both sides

Integrate both sides with respect to $x$: $e^{x}y=\int\frac{e^{x}}{x}dx + C$.

Step5: Solve for $y$

$y = e^{-x}\int\frac{e^{x}}{x}dx+Ce^{-x}$. The integral $\int\frac{e^{x}}{x}dx$ is the exponential integral $\text{Ei}(x)$. So the general solution is $y = e^{-x}\text{Ei}(x)+Ce^{-x}$.

Answer:

$y = e^{-x}\text{Ei}(x)+Ce^{-x}$