Write order of the differential equation of the family of following curves
`Ax+By^2= Bx^2 - Ay`

Answer

To find the order of the differential equation of the family of curves given by the equation \( Ax + By^2 = Bx^2 - Ay \), we will follow these steps:

### Step 1: Rewrite the Given Equation
Start with the original equation:
\[ Ax + By^2 = Bx^2 - Ay \]
Rearranging this gives:
\[ Ax + Ay + By^2 = Bx^2 \]
This can be simplified to:
\[ A(x + y) = Bx^2 - By^2 \]

### Step 2: Isolate Constants
Next, we can express the equation in a more manageable form. Dividing through by \( B \) (assuming \( B \neq 0 \)):
\[ \frac{A}{B}(x + y) = x^2 - y^2 \]
Let \( k = \frac{A}{B} \), then we can write:
\[ k(x + y) = x^2 - y^2 \]

### Step 3: Differentiate the Equation
Now, we will differentiate this equation with respect to \( x \):
\[ \frac{d}{dx}[k(x + y)] = \frac{d}{dx}[x^2 - y^2] \]
Using the product rule on the left side and the chain rule on the right side:
\[ k(1 + \frac{dy}{dx}) = 2x - 2y\frac{dy}{dx} \]

### Step 4: Rearranging the Derivative Equation
Rearranging gives us:
\[ k + k\frac{dy}{dx} = 2x - 2y\frac{dy}{dx} \]
Now, we can isolate \( \frac{dy}{dx} \):
\[ k + k\frac{dy}{dx} + 2y\frac{dy}{dx} = 2x \]
\[ (k + 2y)\frac{dy}{dx} = 2x - k \]
Thus,
\[ \frac{dy}{dx} = \frac{2x - k}{k + 2y} \]

### Step 5: Determine the Order of the Differential Equation
In the expression for \( \frac{dy}{dx} \), we see that the highest derivative present is \( \frac{dy}{dx} \), which is a first derivative. Therefore, the order of the differential equation is:
\[ \text{Order} = 1 \]

### Conclusion
The order of the differential equation corresponding to the family of curves given by \( Ax + By^2 = Bx^2 - Ay \) is **1**.

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