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 $\frac{dy}{dx}=y(xy - 5)$. Rearrange it to the standard Bernoulli form $\frac{dy}{dx}+P(x)y = Q(x)y^n$.
$\frac{dy}{dx}-5y=xy^{2}$. Here $P(x)= - 5$, $Q(x)=x$ and $n = 2$.

Step2: Make a substitution

Let $z=y^{1 - n}=y^{-1}$, then $\frac{dz}{dx}=-y^{-2}\frac{dy}{dx}$. Multiply the differential equation $\frac{dy}{dx}-5y=xy^{2}$ by $-y^{-2}$ to get $-y^{-2}\frac{dy}{dx}+5y^{-1}=-x$. Substituting $z$ and $\frac{dz}{dx}$ gives $\frac{dz}{dx}+5z=-x$.

Step3: Find the integrating - factor

The integrating factor for the linear differential equation $\frac{dz}{dx}+5z=-x$ is $e^{\int5dx}=e^{5x}$.

Step4: Multiply the linear DE by the integrating - factor

We have $e^{5x}\frac{dz}{dx}+5e^{5x}z=-xe^{5x}$. The left - hand side is the derivative of the product $e^{5x}z$ by the product rule, i.e., $\frac{d}{dx}(e^{5x}z)=-xe^{5x}$.

Step5: Integrate both sides

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

Step6: Solve for $z$

$z =-\frac{x}{5}+\frac{1}{25}+Ce^{-5x}$.

Step7: Back - substitute $y$

Since $z = y^{-1}$, we have $\frac{1}{y}=-\frac{x}{5}+\frac{1}{25}+Ce^{-5x}$.
Then $y=\frac{1}{-\frac{x}{5}+\frac{1}{25}+Ce^{-5x}}=\frac{25}{- 5x + 1+25Ce^{-5x}}$.

Answer:

$y=\frac{25}{1 - 5x+25Ce^{-5x}}$