Question

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

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

\[

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

\]
Separate variables: $v^{2}dv=-\frac{1}{x}dx$.

Step3: Integrate both sides

Integrate $\int v^{2}dv=-\int\frac{1}{x}dx$.
Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$) and $\int\frac{1}{x}dx=\ln|x|+C$, we get $\frac{v^{3}}{3}=-\ln|x|+C$.

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

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

Step5: Use the initial condition $y(1) = 3$

Substitute $x = 1$ and $y = 3$ into $\frac{y^{3}}{3x^{3}}=-\ln|x|+C$:
\[
\frac{3^{3}}{3\times1^{3}}=-\ln(1)+C
\]
\[
9 = 0 + C
\]
So $C = 9$.

Step6: Write the final solution

The solution of the initial - value problem is $\frac{y^{3}}{3x^{3}}=-\ln|x| + 9$, or $y^{3}=27x^{3}-3x^{3}\ln|x|$.

Answer:

$y^{3}=27x^{3}-3x^{3}\ln|x|$