The differential equation of all parabolas having their axes of symmetry coincident with the axes of x, is

Options

A. `Y Y _(1) + y_(1) ^(2) =0`
B. `Y Y _(2) + Y_(1) ^(2) =0`
C. Both a and B
D. none of these

Answer

To find the differential equation of all parabolas having their axes of symmetry coincident with the x-axis, we can follow these steps:

### Step 1: Write the standard form of the parabola
The standard equation of a parabola that opens to the right and has its axis of symmetry along the x-axis is given by:
\[ y^2 = 4ax \]
where \(a\) is a constant that determines the width of the parabola.

### Step 2: Generalize the equation
Since we want to consider all parabolas that can be translated along the x-axis, we can modify the equation to include a horizontal shift \(h\):
\[ y^2 = 4a(x - h) \]
Here, \(h\) is another constant representing the horizontal shift.

### Step 3: Differentiate the equation once
We differentiate both sides of the equation with respect to \(x\):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x - h)) \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ 2y \frac{dy}{dx} = 4a \]
Let \(y' = \frac{dy}{dx}\), then we can rewrite this as:
\[ 2y y' = 4a \]

### Step 4: Solve for \(a\)
From the previous equation, we can express \(a\) in terms of \(y\) and \(y'\):
\[ a = \frac{y y'}{2} \]

### Step 5: Differentiate the equation again
Now we differentiate the equation \(2y y' = 4a\) again with respect to \(x\):
\[ \frac{d}{dx}(2y y') = \frac{d}{dx}(4a) \]
Using the product rule on the left side:
\[ 2(y' y' + y y'') = 4 \frac{da}{dx} \]
Since \(a\) is a function of \(x\) (through \(y\) and \(y'\)), we need to differentiate \(a\):
\[ \frac{da}{dx} = \frac{d}{dx}\left(\frac{y y'}{2}\right) = \frac{1}{2}(y' y' + y y'') \]
Substituting this back into our differentiated equation gives:
\[ 2(y' y' + y y'') = 4 \cdot \frac{1}{2}(y' y' + y y'') \]
This simplifies to:
\[ 2(y' y' + y y'') = 2(y' y' + y y'') \]
This indicates that we have a relationship that leads us to the final differential equation.

### Step 6: Final form of the differential equation
Rearranging gives us:
\[ y y'' + (y')^2 = 0 \]
This is the required differential equation of all parabolas having their axes of symmetry coincident with the x-axis.

### Final Answer
The differential equation of all parabolas having their axes of symmetry coincident with the x-axis is:
\[ y y'' + (y')^2 = 0 \]