Respuesta :
Answer: To solve the differential equation \( \frac{{dx}}{{dy}} = 3x^2 - 6x \), we can rearrange it and integrate both sides with respect to \(x\):
\[
\begin{split}
\frac{{dx}}{{dy}} &= 3x^2 - 6x \\
\Rightarrow \frac{{dx}}{{3x^2 - 6x}} &= dy
\end{split}
\]
Now, we integrate both sides:
\[
\begin{split}
\int \frac{{dx}}{{3x^2 - 6x}} &= \int dy \\
\end{split}
\]
The integral on the left side can be simplified by factoring out a common factor in the denominator:
\[
\begin{split}
\int \frac{{dx}}{{3x(x - 2)}} &= \int dy \\
\end{split}
\]
Now, we can perform partial fraction decomposition:
\[
\frac{{1}}{{3x(x - 2)}} = \frac{{A}}{{3x}} + \frac{{B}}{{x - 2}}
\]
Multiplying both sides by \(3x(x - 2)\), we get:
\[
1 = A(x - 2) + 3Bx
\]
Let's find the values of \(A\) and \(B\). Setting \(x = 0\), we get:
\[
1 = -2A \implies A = -\frac{1}{2}
\]
Setting \(x = 2\), we get:
\[
1 = 2B \implies B = \frac{1}{2}
\]
So, our integral becomes:
\[
\begin{split}
\int \left( -\frac{1}{2x} + \frac{1}{2(x - 2)} \right) dx &= \int dy \\
-\frac{1}{2} \int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{x - 2} dx &= \int dy \\
-\frac{1}{2} \ln|x| + \frac{1}{2} \ln|x - 2| &= y + C
\end{split}
\]
Where \(C\) is the constant of integration. Therefore, the solution to the differential equation is:
\[
-\frac{1}{2} \ln|x| + \frac{1}{2} \ln|x - 2| = y + C
\]
