Question

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

Explanation:

Step1: Rewrite the Bernoulli equation

The given differential equation $x^{2}\frac{dy}{dx}-2xy = 6y^{4}$ can be rewritten as $\frac{dy}{dx}-\frac{2}{x}y=\frac{6}{x^{2}}y^{4}$. This is a Bernoulli equation of the form $\frac{dy}{dx}+P(x)y = Q(x)y^{n}$ where $P(x)=-\frac{2}{x}$, $Q(x)=\frac{6}{x^{2}}$ and $n = 4$. We make the substitution $v=y^{1 - n}=y^{- 3}$, then $\frac{dv}{dx}=-3y^{-4}\frac{dy}{dx}$.

Step2: Transform the equation

Multiply the original DE by $y^{-4}$: $y^{-4}\frac{dy}{dx}-\frac{2}{x}y^{-3}=\frac{6}{x^{2}}$. Substituting $v = y^{-3}$ and $\frac{dv}{dx}=-3y^{-4}\frac{dy}{dx}$, we get $-\frac{1}{3}\frac{dv}{dx}-\frac{2}{x}v=\frac{6}{x^{2}}$, or $\frac{dv}{dx}+\frac{6}{x}v=-\frac{18}{x^{2}}$.

Step3: Find the integrating - factor

The integrating factor for the linear differential equation $\frac{dv}{dx}+P(x)v = Q(x)$ (here $P(x)=\frac{6}{x}$) is $\mu(x)=e^{\int\frac{6}{x}dx}=e^{6\ln|x|}=x^{6}$.

Step4: Multiply the transformed equation by the integrating - factor

We have $x^{6}\frac{dv}{dx}+6x^{5}v=- 18x^{4}$. The left - hand side is the derivative of the product $x^{6}v$: $\frac{d}{dx}(x^{6}v)=-18x^{4}$.

Step5: Integrate both sides

Integrating $\frac{d}{dx}(x^{6}v)=-18x^{4}$ with respect to $x$ gives $x^{6}v=-18\times\frac{x^{5}}{5}+C$, or $v =-\frac{18}{5x}+Cx^{-6}$.

Step6: Back - substitute $v = y^{-3}$

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

Step7: Use the initial condition $y(1)=\frac{1}{2}$

When $x = 1$ and $y=\frac{1}{2}$, we have $(\frac{1}{2})^{-3}=-\frac{18}{5\times1}+C\times1^{-6}$. Since $(\frac{1}{2})^{-3}=8$, we get $8=-\frac{18}{5}+C$. Solving for $C$ gives $C = 8+\frac{18}{5}=\frac{40 + 18}{5}=\frac{58}{5}$.

Step8: Write the final solution

$y^{-3}=-\frac{18}{5x}+\frac{58}{5x^{6}}$, or $y^{3}=\frac{5x^{6}}{58 - 18x^{5}}$.

Answer:

$y^{3}=\frac{5x^{6}}{58 - 18x^{5}}$