If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=

1.	If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=2xy\)B. \(x\frac{d^2y}{dx^2}=y_1\)C. \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)D.\(x\frac{d^2y}{dx^2}=2xy\)
Answer» Correct Answer is (C) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\) Given: \(\frac{dy}{dx}=2bx\) \(\frac{d^2y}{dx^2}=2b\neq2xy\) \(x\frac{d^2y}{dx^2}=2bx\) \(=\frac{dy}{dx}\) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y\) \(=2bx-2bx+y\) = y

If y = a + bx2, a, b arbitrary constants, thenA. \(\frac{d^2y}{dx^2}=2xy\)B. \(x\frac{d^2y}{dx^2}=y_1\)C. \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)D.\(x\frac{d^2y}{dx^2}=2xy\)

Answer»

Correct Answer is (C) \(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y=y\)

Given:

\(\frac{dy}{dx}=2bx\)

\(\frac{d^2y}{dx^2}=2b\neq2xy\)

\(x\frac{d^2y}{dx^2}=2bx\)

\(=\frac{dy}{dx}\)

\(x\frac{d^2y}{dx^2}-\frac{dy}{dx}+y\) \(=2bx-2bx+y\)

= y