If $y = e^{ax} \sin(bx)$, prove that $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$.

Question

Accepted Answer

**Step 1: Find the first derivative $\frac{dy}{dx}$ using the product rule**
$$\frac{dy}{dx} = e^{ax}[b \cos(bx)] + \sin(bx)[a e^{ax}] = e^{ax} [b \cos(bx) + a \sin(bx)]$$

**Step 2: Find the second derivative $\frac{d^2y}{dx^2}$**
$$\frac{d^2y}{dx^2} = a e^{ax} [b \cos(bx) + a \sin(bx)] + e^{ax} [-b^2 \sin(bx) + ab \cos(bx)]$$
$$\frac{d^2y}{dx^2} = e^{ax} [ab \cos(bx) + a^2 \sin(bx) - b^2 \sin(bx) + ab \cos(bx)]$$
$$\frac{d^2y}{dx^2} = e^{ax} [2ab \cos(bx) + (a^2 - b^2) \sin(bx)]$$

**Step 3: Substitute into the left-hand side (LHS) of the expression**
$$	ext{LHS} = \frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y$$
$$= e^{ax}[2ab \cos(bx) + (a^2 - b^2) \sin(bx)] - 2a e^{ax}[b \cos(bx) + a \sin(bx)] + (a^2 + b^2) e^{ax} \sin(bx)$$
$$= e^{ax} [ 2ab\cos(bx) + (a^2-b^2)\sin(bx) - 2ab\cos(bx) - 2a^2\sin(bx) + (a^2+b^2)\sin(bx) ]$$

Combining terms inside the bracket:
$$= e^{ax} [ \cos(bx)(2ab - 2ab) + \sin(bx)(a^2 - b^2 - 2a^2 + a^2 + b^2) ] = e^{ax}[0] = 0 = 	ext{RHS}$$

If $y = e^{ax} \sin(bx)$, prove that $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$. - DDCET Q&A

If $y = e^{ax} \sin(bx)$, prove that $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y = 0$.

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