Solution
Given the function:
$$ y = e^{2x} \cos 2x $$
we are tasked with finding the first and second derivatives and proving the given relationships.
(a) Show that $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $
To find $ \frac{dy}{dx} $, we use the product rule, which states that:
$$ \frac{d}{dx} \left( u(x) v(x) \right) = u'(x) v(x) + u(x) v'(x) $$
where $ u(x) = e^{2x} $ and $ v(x) = \cos 2x $.
First, differentiate $ u(x) = e^{2x} $:
$$ u'(x) = \frac{d}{dx} e^{2x} = 2e^{2x} $$
Next, differentiate $ v(x) = \cos 2x $:
$$ v'(x) = \frac{d}{dx} \cos 2x = -2 \sin 2x $$
Now apply the product rule:
$$ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 2e^{2x} \cos 2x - e^{2x} \cdot 2 \sin 2x $$
$$ \frac{dy}{dx} = 2e^{2x} \cos 2x - 2e^{2x} \sin 2x $$
Since $ y = e^{2x} \cos 2x $, we can rewrite this as:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
Thus, we have shown that:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
(b) Hence show that $ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $
To find $ \frac{d^2y}{dx^2} $, differentiate $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $ again using the chain rule and product rule.
We have:
$$ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $$
Differentiate both terms:
1. The derivative of $ 2y $ is:
$$ \frac{d}{dx} (2y) = 2 \frac{dy}{dx} $$
2. The derivative of $ -2e^{2x} \sin 2x $ is found using the product rule. Let $ u(x) = -2e^{2x} $ and $ v(x) = \sin 2x $. First, differentiate $ u(x) $ and $ v(x) $:
$$ u'(x) = -4e^{2x}, \quad v'(x) = 2 \cos 2x $$
Apply the product rule:
$$ \frac{d}{dx} (-2e^{2x} \sin 2x) = -4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Now, combine the derivatives:
$$ \frac{d^2y}{dx^2} = 2 \frac{dy}{dx} - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Substitute $ \frac{dy}{dx} = 2y - 2e^{2x} \sin 2x $ from part (a):
$$ \frac{d^2y}{dx^2} = 2(2y - 2e^{2x} \sin 2x) - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
$$ \frac{d^2y}{dx^2} = 4y - 4e^{2x} \sin 2x - 4e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
$$ \frac{d^2y}{dx^2} = 4y - 8e^{2x} \sin 2x - 4e^{2x} \cos 2x $$
Finally, notice that:
$$ 4y = 4e^{2x} \cos 2x $$
so we have:
$$ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $$
Thus, we have shown that:
$$ \frac{d^2y}{dx^2} = 4 \frac{dy}{dx} - 8y $$
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