Question

solve the given initial - value problem. the de is homogeneous.
$\frac{dy}{dx}=\frac{xy^{2}}{y^{3}-x^{3}},y(1) = 4$

Explanation:

Step1: Rewrite the differential equation

Given $\frac{dy}{dx}=\frac{y^{3}-x^{3}}{xy^{2}}$. Let $y = vx$, then $\frac{dy}{dx}=v + x\frac{dv}{dx}$. Substitute $y = vx$ into the differential equation:
\[

$$\begin{align*} v + x\frac{dv}{dx}&=\frac{(vx)^{3}-x^{3}}{x(vx)^{2}}\\ v + x\frac{dv}{dx}&=\frac{v^{3}x^{3}-x^{3}}{v^{2}x^{3}}\\ v + x\frac{dv}{dx}&=\frac{v^{3}-1}{v^{2}} \end{align*}$$

\]

Step2: Simplify and separate variables

\[

$$\begin{align*} x\frac{dv}{dx}&=\frac{v^{3}-1}{v^{2}}-v\\ x\frac{dv}{dx}&=\frac{v^{3}-1 - v^{3}}{v^{2}}\\ x\frac{dv}{dx}&=-\frac{1}{v^{2}}\\ v^{2}dv&=-\frac{1}{x}dx \end{align*}$$

\]

Step3: Integrate both sides

Integrate $\int v^{2}dv=-\int\frac{1}{x}dx$.
We know that $\int v^{2}dv=\frac{v^{3}}{3}+C_1$ and $\int\frac{1}{x}dx=\ln|x|+C_2$. So $\frac{v^{3}}{3}=-\ln|x| + C$.

Step4: Substitute back $v=\frac{y}{x}$

\[
\frac{y^{3}}{3x^{3}}=-\ln|x| + C
\]

Step5: Use the initial - condition $y(1) = 4$

When $x = 1$ and $y = 4$, we have $\frac{4^{3}}{3\times1^{3}}=-\ln(1)+C$. Since $\ln(1)=0$, then $C=\frac{64}{3}$.

Step6: Write the final solution

The solution of the initial - value problem is $\frac{y^{3}}{3x^{3}}=-\ln|x|+\frac{64}{3}$, or $y^{3}=- 3x^{3}\ln|x| + 64x^{3}$.

Answer:

$y^{3}=-3x^{3}\ln|x| + 64x^{3}$