Question

solve the given initial - value problem.
$\frac{dy}{dx}-2xy = 4y^{4}$, $y(1)=\frac{1}{2}$

Explanation:

Step1: Rewrite the differential equation

The given differential equation $\frac{dy}{dx}- \frac{2y}{x}=4y^{4}$ is a Bernoulli - type differential equation. Divide through by $y^{4}$ to get $y^{-4}\frac{dy}{dx}-2xy^{-3}=4$. Let $v = y^{-3}$, then $\frac{dv}{dx}=- 3y^{-4}\frac{dy}{dx}$, and $y^{-4}\frac{dy}{dx}=-\frac{1}{3}\frac{dv}{dx}$. The equation becomes $-\frac{1}{3}\frac{dv}{dx}-\frac{2}{x}v = 4$, or $\frac{dv}{dx}+\frac{6}{x}v=-12$.

Step2: Find the integrating factor

The integrating factor for the linear - first - order differential equation $\frac{dv}{dx}+P(x)v = Q(x)$ (here $P(x)=\frac{6}{x}$ and $Q(x)=-12$) is $\mu(x)=e^{\int P(x)dx}$. Calculate $\int\frac{6}{x}dx = 6\ln|x|=\ln(x^{6})$ (assuming $x>0$), so $\mu(x)=x^{6}$.

Step3: Multiply the linear equation by the integrating factor

Multiply $\frac{dv}{dx}+\frac{6}{x}v=-12$ by $x^{6}$: $x^{6}\frac{dv}{dx}+6x^{5}v=-12x^{6}$. The left - hand side is the derivative of the product $x^{6}v$ by the product rule, i.e., $\frac{d}{dx}(x^{6}v)=-12x^{6}$.

Step4: Integrate both sides

Integrate $\frac{d}{dx}(x^{6}v)=-12x^{6}$ with respect to $x$: $x^{6}v=\int - 12x^{6}dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$), we have $\int - 12x^{6}dx=-12\times\frac{x^{7}}{7}+C=-\frac{12x^{7}}{7}+C$. So $v =-\frac{12x}{7}+\frac{C}{x^{6}}$.

Step5: Substitute back $v = y^{-3}$

Since $v = y^{-3}$, we have $y^{-3}=-\frac{12x}{7}+\frac{C}{x^{6}}$.

Step6: Use the initial condition $y(1)=\frac{1}{2}$

When $x = 1$ and $y=\frac{1}{2}$, $(\frac{1}{2})^{-3}=-\frac{12\times1}{7}+\frac{C}{1^{6}}$. Since $(\frac{1}{2})^{-3}=8$, we get $8=-\frac{12}{7}+C$. Solving for $C$, we have $C = 8+\frac{12}{7}=\frac{56 + 12}{7}=\frac{68}{7}$.

Step7: Write the final solution

$y^{-3}=-\frac{12x}{7}+\frac{68}{7x^{6}}$, or $y^{3}=\frac{7x^{6}}{68 - 12x^{7}}$.

Answer:

$y^{3}=\frac{7x^{6}}{68 - 12x^{7}}$