Question

Find the differential equation which has $y = Ax^2 + Bx$.

• A. $y = x \frac{dy}{dx} - x^2 \frac{d^2y}{dx^2}$
• B. $y = x \frac{dy}{dx} - \frac{1}{2} x^2 \frac{d^2y}{dx^2}$
• C. $y = x \frac{dy}{dx} + x^2 \frac{d^2y}{dx^2}$
• D. $y = x \frac{dy}{dx} + \frac{1}{2} x^2 \frac{d^2y}{dx^2}$

Answer 1

Answer: (B)

$y = x \frac{dy}{dx} - \frac{1}{2} x^2 \frac{d^2y}{dx^2}$

Answer 2

To find the differential equation from the given general solution $y = Ax^2 + Bx$, we need to eliminate the arbitrary constants A and B. Since there are two arbitrary constants, the resulting differential equation will be of the second order.

Given equation: $y = Ax^2 + Bx \quad \cdots (1)$

Differentiate equation (1) with respect to x to find the first derivative: $\frac{dy}{dx} = 2Ax + B \quad \cdots (2)$

Differentiate equation (2) with respect to x to find the second derivative: $\frac{d^2y}{dx^2} = 2A \quad \cdots (3)$

Now, we need to eliminate A and B using equations (1), (2), and (3).

From equation (3), we can express A in terms of $\frac{d^2y}{dx^2}$: $A = \frac{1}{2} \frac{d^2y}{dx^2}$

Substitute this expression for A into equation (2): $\frac{dy}{dx} = 2 \left(\frac{1}{2} \frac{d^2y}{dx^2}\right) x + B$ $\frac{dy}{dx} = x \frac{d^2y}{dx^2} + B$

Now, express B in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$: $B = \frac{dy}{dx} - x \frac{d^2y}{dx^2}$

Finally, substitute the expressions for A and B back into the original equation (1): $y = \left(\frac{1}{2} \frac{d^2y}{dx^2}\right) x^2 + \left(\frac{dy}{dx} - x \frac{d^2y}{dx^2}\right) x$ $y = \frac{1}{2} x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} - x^2 \frac{d^2y}{dx^2}$

Combine the terms containing $\frac{d^2y}{dx^2}$: $y = x \frac{dy}{dx} + \left(\frac{1}{2} - 1\right) x^2 \frac{d^2y}{dx^2}$ $y = x \frac{dy}{dx} - \frac{1}{2} x^2 \frac{d^2y}{dx^2}$

This is the required differential equation.