Question

The equation \(\dfrac{d^2y}{dx^2} - y = 0\) has a solution of the form \(y = e^{Ax}\). The value(s) of \(A\) satisfying this is/are:

• A. \(0\)
• B. \(1\)
• C. \(-1\)
• D. \(-\infty\)

Hint:

Substitute \(y=e^{Ax}\) into the equation to get the characteristic equation \(A^2-1=0\), then solve for A.

Answer 1

The Correct Option is B, C

This equation is a standard second-order linear differential equation with constant coefficients. Instead of substituting each option one by one, it helps to first build the characteristic equation directly, then check the options against it.

1. Setting up the characteristic equation: Replacing $\dfrac{d^2y}{dx^2}$ with $A^2$ and $y$ with $1$ (since $y = e^{Ax}$ and each derivative brings down a factor of $A$), the equation $\dfrac{d^2y}{dx^2} - y = 0$ becomes the algebraic equation $A^2 - 1 = 0$.
2. Option 0: Plugging $A=0$ gives $0^2 - 1 = -1$, which is not zero, so this value does not solve the equation.
3. Option 1: Plugging $A=1$ gives $1^2 - 1 = 0$. This satisfies the characteristic equation.
4. Option -1: Plugging $A=-1$ gives $(-1)^2 - 1 = 1 - 1 = 0$. This also satisfies the characteristic equation.
5. Option $-\infty$: This is not a finite number, so it cannot be a root of the polynomial $A^2 - 1 = 0$, which by the fundamental theorem of algebra has exactly two finite roots.

Since $A^2 - 1 = (A-1)(A+1) = 0$ has roots $A = 1$ and $A = -1$, and these are exactly the two finite values among the four options that work, both must be selected.

Let's summarize:

• Substituting $y = e^{Ax}$ into a linear constant-coefficient equation always turns it into a polynomial in $A$, the characteristic equation.
• Here that polynomial is $A^2 - 1 = 0$, giving the two roots $A = \pm 1$.

The values of $A$ that satisfy the equation are $1$ and $-1$.