Find a general solution. Show the steps of derivation. Check your answer by substitution. y³y'+x³=0

$y^{3}y'+x^{3}=0\;\;\; \Rightarrow \;\;\; \textbf{Simplify}$

$\dfrac{dy}{dx}=\dfrac{-x^{3}}{y^{3}}\;\;\; \Rightarrow \;\;\; \textbf{Put(y) on LHS and (x) on RHS}$

$y^{3}dy=-x^{3}dx\;\;\; \Rightarrow \;\;\; \textbf{Integration on both sides}$

$\int y^{3}dy=-\int x^{3}dx$

$\dfrac{y^{4}}{4}+C_{1}=\dfrac{-x^{4}}{4}+C_{2}\;\;\; \Rightarrow \;\;\; \textbf{Simplify}$

$$\boxed{y^{4}+x^{4}=C}$$

The ODE $y^3y' + x^3 = 0$ is $\textbf{separable}$ because it can be written:

$$\begin{align*} y^3y' = -x^3 \iff y^3 \frac{dy}{dx} = -x^3 \iff y^3 dy = -x^3 dx \end{align*}$$

Now, the variables are separated, x appears only on the right side, and y only on the left.

Integrate the left side in relation to $y$, and the right side in relation to $x$:

$$\begin{align*} \int y^3 dy &= \int \left(-x^3 \right) dx \\ \frac{y^4}{4} &= -\frac{x^4}{4} + c \end{align*}$$

Next, we need to isolate y:

$$y^4 = 4c - x^4 \implies y^4 = C - x^4 \implies \boxed{\textcolor{Fuchsia}{y = \sqrt[4]{C - x^4}}}$$

$$\begin{align*} \implies &y' =\overset{\textcolor{Fuchsia}{\text{Power Rule}}}{ \frac{1}{4} (C - x^4)^{-\frac{3}{4}}} \overset{\textcolor{Fuchsia}{\text{Chain Rule}}}{\cdot} (C - x^4)' \\ & = \frac{1}{4} (C - x^4)^{-\frac{3}{4}} (-4 x^3) \\ & = -\frac{x^3}{(C - x^4)^{\frac{3}{4}}} \\ &= - \frac{x^3}{(\sqrt[4]{C - x^4})^3} \end{align*}$$

Now, we can substitute $y'$ and $y$ in the equation:

$$y^3y' + x^3 = 0 \iff \cancel{(\sqrt[4]{C - x^4})^3} \cdot \left( - \frac{x^3}{(\cancel{\sqrt[4]{C - x^4})^3}}\right) + x^3 = 0 \iff -x^3 + x^3 = 0$$

This way, we verify that $y = \sqrt[4]{C - x^4}$ is the solution of the ODE.