Question

solve the given initial - value problem. the de is a bernoulli equation.
$x^{2}\frac{dy}{dx}-2xy = 6y^{4}$, $y(1)=\frac{1}{4}$

Explanation:

Step1: Rewrite the Bernoulli equation

The given Bernoulli equation $x^{2}\frac{dy}{dx}-2xy = 6y^{4}$ can be rewritten in the standard form $\frac{dy}{dx}-\frac{2}{x}y=\frac{6}{x^{2}}y^{4}$. Let $z = y^{- 3}$, then $\frac{dz}{dx}=-3y^{-4}\frac{dy}{dx}$. Multiply the original equation by $y^{-4}$: $y^{-4}\frac{dy}{dx}-\frac{2}{x}y^{-3}=\frac{6}{x^{2}}$. Substituting $z = y^{-3}$ and $\frac{dz}{dx}=-3y^{-4}\frac{dy}{dx}$, we get $-\frac{1}{3}\frac{dz}{dx}-\frac{2}{x}z=\frac{6}{x^{2}}$, or $\frac{dz}{dx}+\frac{6}{x}z =-\frac{18}{x^{2}}$.

Step2: Find the integrating factor

The integrating factor for the linear - first - order differential equation $\frac{dz}{dx}+P(x)z = Q(x)$ (here $P(x)=\frac{6}{x}$ and $Q(x)=-\frac{18}{x^{2}}$) is $\mu(x)=e^{\int P(x)dx}=e^{\int\frac{6}{x}dx}=e^{6\ln|x|}=x^{6}$.

Step3: Multiply the linear equation by the integrating factor

Multiply $\frac{dz}{dx}+\frac{6}{x}z =-\frac{18}{x^{2}}$ by $x^{6}$: $x^{6}\frac{dz}{dx}+6x^{5}z=- 18x^{4}$. The left - hand side is the derivative of the product $x^{6}z$ with respect to $x$, i.e., $\frac{d}{dx}(x^{6}z)=-18x^{4}$.

Step4: Integrate both sides

Integrate $\frac{d}{dx}(x^{6}z)=-18x^{4}$ with respect to $x$: $x^{6}z=\int-18x^{4}dx=-18\times\frac{x^{5}}{5}+C=-\frac{18}{5}x^{5}+C$. So, $z =-\frac{18}{5x}+ \frac{C}{x^{6}}$.

Step5: Substitute back $y$

Since $z = y^{-3}$, we have $y^{-3}=-\frac{18}{5x}+\frac{C}{x^{6}}$.

Step6: Use the initial condition

Given $y(1)=\frac{1}{4}$, when $x = 1$ and $y=\frac{1}{4}$, then $(\frac{1}{4})^{-3}=-\frac{18}{5\times1}+\frac{C}{1^{6}}$. $64=-\frac{18}{5}+C$, so $C = 64+\frac{18}{5}=\frac{320 + 18}{5}=\frac{338}{5}$.

Step7: Write the final solution

$y^{-3}=-\frac{18}{5x}+\frac{338}{5x^{6}}=\frac{-18x^{5}+338}{5x^{6}}$, and $y^{3}=\frac{5x^{6}}{338 - 18x^{5}}$.

Answer:

$y^{3}=\frac{5x^{6}}{338 - 18x^{5}}$