Answer:
[tex]y = x^3 +C[/tex]
Explanation:
Given:
[tex]\text{Solve,} -6y'+18x^2=0.[/tex]
[tex]\Longrightarrow -6y'+18x^2=0\\\\\Longrightarrow [-6y'+18x^2=0]-\frac{1}{6} \\\\\Longrightarrow y'-3x^2=0\\\\\Longrightarrow \frac{dy}{dx} =3x^2\\[/tex]
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Using Speration of Varibles:}}\\\frac{dy}{dx}=f(x)g(y) \\\Rightarrow \int\frac{dy}{g(y)}=\int f(x)gx \end{array}\right}[/tex]
[tex]\Longrightarrow \frac{dy}{dx} =3x^2\\\\\Longrightarrow dy =3x^2dx\\\\\Longrightarrow \int dy =\int 3x^2dx\\\\\Longrightarrow \boxed{\boxed{y = x^3 +C}} \therefore Sol.[/tex]
Thus, the given first-order differential equation is solved.
