Problem and Solution
Given that $$ y = e^{3x} \sin 2x $$ show that $$ 13y + \frac{d^2 y}{dx^2} = 6 \frac{dy}{dx} $$
Step 1: Find $ \frac{dy}{dx} $
We will differentiate $$ y = e^{3x} \sin 2x $$ using the product rule. Recall that $$ \frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v' $$ where $$ u = e^{3x} \quad \text{and} \quad v = \sin 2x $$
First, compute the derivatives:
$$ u' = \frac{d}{dx} e^{3x} = 3e^{3x}, \quad v' = \frac{d}{dx} \sin 2x = 2\cos 2x $$
Now, apply the product rule:
$$ \frac{dy}{dx} = 3e^{3x} \sin 2x + e^{3x} \cdot 2\cos 2x $$
Thus, we have:
$$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$
Step 2: Find $ \frac{d^2 y}{dx^2} $
Next, differentiate $$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$ using the product rule again.
Let $$ u = e^{3x} \quad \text{and} \quad v = 3\sin 2x + 2\cos 2x $$ so
$$ u' = 3e^{3x}, \quad v' = 3 \cdot 2\cos 2x - 2 \cdot 2\sin 2x = 6\cos 2x - 4\sin 2x $$
Now, apply the product rule:
$$ \frac{d^2 y}{dx^2} = 3e^{3x}(3\sin 2x + 2\cos 2x) + e^{3x}(6\cos 2x - 4\sin 2x) $$
Simplify:
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ 9\sin 2x + 6\cos 2x + 6\cos 2x - 4\sin 2x \right] $$
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ (9 - 4)\sin 2x + (6 + 6)\cos 2x \right] $$
$$ \frac{d^2 y}{dx^2} = e^{3x} \left[ 5\sin 2x + 12\cos 2x \right] $$
Step 3: Verify the equation $ 13y + \frac{d^2 y}{dx^2} = 6 \frac{dy}{dx} $
Now substitute $$ y = e^{3x}\sin 2x $$ $$ \frac{dy}{dx} = e^{3x}(3\sin 2x + 2\cos 2x) $$ and $$ \frac{d^2 y}{dx^2} = e^{3x}(5\sin 2x + 12\cos 2x) $$ into the equation.
The left-hand side:
$$ 13y + \frac{d^2 y}{dx^2} = 13e^{3x}\sin 2x + e^{3x}(5\sin 2x + 12\cos 2x) $$
Factor out $e^{3x}$:
$$ 13e^{3x}\sin 2x + e^{3x}(5\sin 2x + 12\cos 2x) = e^{3x} \left[ 13\sin 2x + 5\sin 2x + 12\cos 2x \right] $$
$$ = e^{3x} \left[ (13 + 5)\sin 2x + 12\cos 2x \right] = e^{3x} \left[ 18\sin 2x + 12\cos 2x \right] $$
The right-hand side:
$$ 6\frac{dy}{dx} = 6e^{3x}(3\sin 2x + 2\cos 2x) $$
$$ = e^{3x} \left[ 18\sin 2x + 12\cos 2x \right] $$
Thus, both sides are equal:
$$ 13y + \frac{d^2 y}{dx^2} = 6\frac{dy}{dx} $$
Final Answer
We have shown that
$$ 13y + \frac{d^2 y}{dx^2} = 6\frac{dy}{dx} $$
Post a Comment