Question

solve the given differential equation by using an appropriate substitution. the de is a bernoulli equation.
$y\frac{dx}{dy}+x = \frac{1}{y^{2}}$

Explanation:

Step1: Rewrite the Bernoulli equation

The given differential equation $\frac{dy}{dx}+y = \frac{1}{y^{2}}$ can be rewritten as $\frac{dy}{dx}+y=y^{- 2}$. Multiply through by $y^{2}$ to get $y^{2}\frac{dy}{dx}+y^{3}=1$. Let $u = y^{3}$, then $\frac{du}{dx}=3y^{2}\frac{dy}{dx}$, and $y^{2}\frac{dy}{dx}=\frac{1}{3}\frac{du}{dx}$.

Step2: Substitute into the equation

Substituting $u = y^{3}$ and $y^{2}\frac{dy}{dx}=\frac{1}{3}\frac{du}{dx}$ into $y^{2}\frac{dy}{dx}+y^{3}=1$, we obtain $\frac{1}{3}\frac{du}{dx}+u = 1$, or $\frac{du}{dx}+3u = 3$.

Step3: Find the integrating - factor

The integrating factor for the linear differential equation $\frac{du}{dx}+3u = 3$ is $e^{\int3dx}=e^{3x}$.

Step4: Multiply the equation by the integrating - factor

Multiply $\frac{du}{dx}+3u = 3$ by $e^{3x}$: $e^{3x}\frac{du}{dx}+3e^{3x}u = 3e^{3x}$. The left - hand side is the derivative of the product $e^{3x}u$ by the product rule, i.e., $\frac{d}{dx}(e^{3x}u)=3e^{3x}$.

Step5: Integrate both sides

Integrate $\frac{d}{dx}(e^{3x}u)=3e^{3x}$ with respect to $x$. $\int\frac{d}{dx}(e^{3x}u)dx=\int3e^{3x}dx$. So $e^{3x}u = e^{3x}+C$, where $C$ is the constant of integration.

Step6: Solve for $u$

Divide both sides of $e^{3x}u = e^{3x}+C$ by $e^{3x}$ to get $u = 1 + Ce^{-3x}$.

Step7: Substitute back $y$

Since $u = y^{3}$, we have $y^{3}=1 + Ce^{-3x}$.

Answer:

$y^{3}=1 + Ce^{-3x}$