To solve the problem, we need to find the curve that passes through the point (0,0) and has a given slope of the tangent at any point (x,y). The slope is given by the expression:
\[
\frac{dy}{dx} = \frac{x^2 - 4x + y + 8}{x - 2}
\]
### Step 1: Rearranging the Differential Equation
We can rearrange the equation to isolate \(y\):
\[
\frac{dy}{dx} - \frac{1}{x - 2} y = \frac{x^2 - 4x + 8}{x - 2}
\]
This is a linear first-order differential equation in the standard form:
\[
\frac{dy}{dx} + P(x)y = Q(x)
\]
where \(P(x) = -\frac{1}{x - 2}\) and \(Q(x) = \frac{x^2 - 4x + 8}{x - 2}\).
### Step 2: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:
\[
I(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x - 2} \, dx} = e^{-\ln|x - 2|} = \frac{1}{|x - 2|}
\]
For our purposes, we can use \(I(x) = \frac{1}{x - 2}\).
### Step 3: Multiplying the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:
\[
\frac{1}{x - 2} \frac{dy}{dx} - \frac{1}{(x - 2)^2} y = \frac{x^2 - 4x + 8}{(x - 2)^2}
\]
### Step 4: Integrating Both Sides
The left-hand side can be simplified to:
\[
\frac{d}{dx}\left(\frac{y}{x - 2}\right) = \frac{x^2 - 4x + 8}{(x - 2)^2}
\]
Now we integrate both sides:
\[
\int \frac{d}{dx}\left(\frac{y}{x - 2}\right) \, dx = \int \frac{x^2 - 4x + 8}{(x - 2)^2} \, dx
\]
The left side gives:
\[
\frac{y}{x - 2}
\]
For the right side, we can perform polynomial long division:
\[
\frac{x^2 - 4x + 8}{(x - 2)^2} = 1 + \frac{4}{(x - 2)^2}
\]
Thus, we have:
\[
\frac{y}{x - 2} = x + 4 \cdot \int \frac{1}{(x - 2)^2} \, dx + C
\]
The integral of \(\frac{1}{(x - 2)^2}\) is \(-\frac{1}{x - 2}\):
\[
\frac{y}{x - 2} = x - \frac{4}{x - 2} + C
\]
### Step 5: Solving for \(y\)
Multiplying through by \(x - 2\):
\[
y = (x - 2)(x - \frac{4}{x - 2} + C)
\]
### Step 6: Applying the Initial Condition
Since the curve passes through the point (0,0):
\[
0 = (0 - 2)(0 - 4 + C) \Rightarrow 0 = -2(-4 + C)
\]
This implies:
\[
-4 + C = 0 \Rightarrow C = 4
\]
### Final Equation of the Curve
Substituting \(C\) back into the equation:
\[
y = (x - 2)(x - 4 + 4) = (x - 2)(x) = x^2 - 2x
\]
### Step 7: Finding Another Point on the Curve
To find another point that the curve passes through, we can check various points. We can substitute values into the equation \(y = x^2 - 2x\) to see which points satisfy it.
1. For \(x = 4\):
\[
y = 4^2 - 2 \cdot 4 = 16 - 8 = 8 \quad \text{(Point: (4, 8))}
\]
2. For \(x = 5\):
\[
y = 5^2 - 2 \cdot 5 = 25 - 10 = 15 \quad \text{(Point: (5, 15))}
\]
3. For \(x = 3\):
\[
y = 3^2 - 2 \cdot 3 = 9 - 6 = 3 \quad \text{(Point: (3, 3))}
\]
### Conclusion
The curve passes through the point (4, 8) and (5, 15) among others.
