Question

solve the given differential equation by using an appropriate substitution. the de is a bernoulli equation.
\\( \frac{dy}{dx}=y(xy^{5}-1) \\)

Explanation:

Step1: Rewrite the Bernoulli equation

The given differential equation is $\frac{dy}{dx}=y(xy^5 - 1)=xy^{6}-y$. Divide by $y^{6}$ to get $y^{-6}\frac{dy}{dx}-y^{-5}=-x$. Let $u = y^{-5}$, then $\frac{du}{dx}=- 5y^{-6}\frac{dy}{dx}$, and $y^{-6}\frac{dy}{dx}=-\frac{1}{5}\frac{du}{dx}$.

Step2: Substitute into the equation

Substituting $u = y^{-5}$ and $y^{-6}\frac{dy}{dx}=-\frac{1}{5}\frac{du}{dx}$ into the equation $y^{-6}\frac{dy}{dx}-y^{-5}=-x$, we obtain $-\frac{1}{5}\frac{du}{dx}-u=-x$, or $\frac{du}{dx}+5u = 5x$.

Step3: Find the integrating - factor

The integrating factor for the linear differential equation $\frac{du}{dx}+P(x)u = Q(x)$ (here $P(x) = 5$ and $Q(x)=5x$) is $e^{\int P(x)dx}=e^{\int5dx}=e^{5x}$.

Step4: Multiply the equation by the integrating - factor

Multiply $\frac{du}{dx}+5u = 5x$ by $e^{5x}$: $e^{5x}\frac{du}{dx}+5e^{5x}u = 5xe^{5x}$. The left - hand side is the derivative of the product $e^{5x}u$ by the product rule, so $\frac{d}{dx}(e^{5x}u)=5xe^{5x}$.

Step5: Integrate both sides

Integrate $\frac{d}{dx}(e^{5x}u)=5xe^{5x}$ with respect to $x$. We use integration by parts for $\int xe^{5x}dx$. Let $v = x$, $dw=e^{5x}dx$, then $dv = dx$, $w=\frac{1}{5}e^{5x}$. So $\int xe^{5x}dx=\frac{1}{5}xe^{5x}-\frac{1}{25}e^{5x}+C$. Then $e^{5x}u = 5(\frac{1}{5}xe^{5x}-\frac{1}{25}e^{5x})+C=xe^{5x}-\frac{1}{5}e^{5x}+C$.

Step6: Solve for $u$

$u = x-\frac{1}{5}+Ce^{-5x}$.

Step7: Substitute back for $y$

Since $u = y^{-5}$, we have $y^{-5}=x-\frac{1}{5}+Ce^{-5x}$, or $y^{5}=\frac{1}{x - \frac{1}{5}+Ce^{-5x}}$.

Answer:

$y^{5}=\frac{1}{x-\frac{1}{5}+Ce^{-5x}}$