Question

using separation of variables, solve the differential equation,
\\(\left(8 + x^{4}\
ight)\frac{dy}{dx} = \frac{x^{3}}{y}\\).
use \\(c\\) to represent the arbitrary constant.
\\(y^{2} = \square\\)

Explanation:

Step1: Separate Variables

Given the differential equation \((8 + x^4)\frac{dy}{dx}=\frac{x^3}{y}\), we multiply both sides by \(y\) and \((8 + x^4)^{-1}dx\) to separate variables:
\(y\ dy=\frac{x^3}{8 + x^4}dx\)

Step2: Integrate Both Sides

Integrate the left - hand side with respect to \(y\) and the right - hand side with respect to \(x\).

• For the left - hand side: \(\int y\ dy=\frac{y^{2}}{2}+C_1\) (where \(C_1\) is the constant of integration for the left integral).
• For the right - hand side: Let \(u = 8+x^4\), then \(du = 4x^{3}dx\), or \(\frac{1}{4}du=x^{3}dx\). So \(\int\frac{x^3}{8 + x^4}dx=\frac{1}{4}\int\frac{du}{u}=\frac{1}{4}\ln|u|+C_2=\frac{1}{4}\ln(8 + x^4)+C_2\) (since \(8 + x^4>0\) for all real \(x\), we can drop the absolute value).

Combining the two integrals (and combining the constants \(C_1\) and \(C_2\) into a single constant \(C\)):
\(\frac{y^{2}}{2}=\frac{1}{4}\ln(8 + x^4)+C\)

Step3: Solve for \(y^{2}\)

Multiply both sides of the equation by \(2\) to get:
\(y^{2}=\frac{1}{2}\ln(8 + x^4)+2C\)
We can rewrite \(2C\) as just \(C\) (since \(C\) is an arbitrary constant) to obtain:
\(y^{2}=\frac{1}{2}\ln(8 + x^4)+C\)

Answer:

\(\frac{1}{2}\ln(8 + x^4)+C\) (or equivalent forms with the constant re - labeled)