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Sagot :
Certainly! Let's tackle the problem in two parts: proving the given differential equation for [tex]\( y \)[/tex] and then using that result to prove the analogous differential equation for [tex]\( z \)[/tex].
### Part 1: Proving [tex]\( \frac{d^2 y}{dx^2} + 36y = 18 \)[/tex] for [tex]\( y = \sin^2(3x) \)[/tex]
1. Define [tex]\( y \)[/tex]:
[tex]\[ y = \sin^2(3x) \][/tex]
2. Find the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx} [\sin^2(3x)] = 2 \sin(3x) \cdot \cos(3x) \cdot 3 = 6 \sin(3x) \cos(3x) \][/tex]
3. Simplify using a trigonometric identity:
Recall the double-angle identity: [tex]\( \sin(2A) = 2 \sin(A) \cos(A) \)[/tex]. Thus:
[tex]\[ 6 \sin(3x) \cos(3x) = 3 \sin(6x) \][/tex]
So:
[tex]\[ \frac{dy}{dx} = 3 \sin(6x) \][/tex]
4. Find the second derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( 3 \sin(6x) \right) = 3 \cdot 6 \cos(6x) = 18 \cos(6x) \][/tex]
5. Construct and simplify the expression [tex]\( \frac{d^2 y}{dx^2} + 36y \)[/tex]:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \cos(6x) + 36 \sin^2(3x) \][/tex]
6. Simplify the expression further:
Using the identity [tex]\( \sin^2(A) = \frac{1 - \cos(2A)}{2} \)[/tex]:
[tex]\[ \sin^2(3x) = \frac{1 - \cos(6x)}{2} \][/tex]
Therefore:
[tex]\[ 36 \sin^2(3x) = 36 \cdot \frac{1 - \cos(6x)}{2} = 18 - 18 \cos(6x) \][/tex]
7. Combine the terms:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \cos(6x) + 18 - 18 \cos(6x) = 18 \][/tex]
Thus, we have shown:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \][/tex]
### Part 2: Proving [tex]\( \frac{d^2 z}{dx^2} + 36z = 18 \)[/tex] for [tex]\( z = \cos^2(3x) \)[/tex]
1. Define [tex]\( z \)[/tex]:
[tex]\[ z = \cos^2(3x) \][/tex]
2. Find the first derivative of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule:
[tex]\[ \frac{dz}{dx} = \frac{d}{dx} [\cos^2(3x)] = 2 \cos(3x) \cdot (-\sin(3x)) \cdot 3 = -6 \sin(3x) \cos(3x) \][/tex]
3. Simplify using a trigonometric identity:
Recall again the double-angle identity:
[tex]\[ -6 \sin(3x) \cos(3x) = -3 \sin(6x) \][/tex]
So:
[tex]\[ \frac{dz}{dx} = -3 \sin(6x) \][/tex]
4. Find the second derivative of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d^2 z}{dx^2} = \frac{d}{dx} \left( -3 \sin(6x) \right) = -3 \cdot 6 \cos(6x) = -18 \cos(6x) \][/tex]
5. Construct and simplify the expression [tex]\( \frac{d^2 z}{dx^2} + 36z \)[/tex]:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = -18 \cos(6x) + 36 \cos^2(3x) \][/tex]
6. Simplify the expression further:
Using the identity [tex]\( \cos^2(A) = \frac{1 + \cos(2A)}{2} \)[/tex]:
[tex]\[ \cos^2(3x) = \frac{1 + \cos(6x)}{2} \][/tex]
Therefore:
[tex]\[ 36 \cos^2(3x) = 36 \cdot \frac{1 + \cos(6x)}{2} = 18 + 18 \cos(6x) \][/tex]
7. Combine the terms:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = -18 \cos(6x) + 18 + 18 \cos(6x) = 18 \][/tex]
Thus, we have shown:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = 18 \][/tex]
Hence, both differential equations are satisfactorily proven.
### Part 1: Proving [tex]\( \frac{d^2 y}{dx^2} + 36y = 18 \)[/tex] for [tex]\( y = \sin^2(3x) \)[/tex]
1. Define [tex]\( y \)[/tex]:
[tex]\[ y = \sin^2(3x) \][/tex]
2. Find the first derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx} [\sin^2(3x)] = 2 \sin(3x) \cdot \cos(3x) \cdot 3 = 6 \sin(3x) \cos(3x) \][/tex]
3. Simplify using a trigonometric identity:
Recall the double-angle identity: [tex]\( \sin(2A) = 2 \sin(A) \cos(A) \)[/tex]. Thus:
[tex]\[ 6 \sin(3x) \cos(3x) = 3 \sin(6x) \][/tex]
So:
[tex]\[ \frac{dy}{dx} = 3 \sin(6x) \][/tex]
4. Find the second derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( 3 \sin(6x) \right) = 3 \cdot 6 \cos(6x) = 18 \cos(6x) \][/tex]
5. Construct and simplify the expression [tex]\( \frac{d^2 y}{dx^2} + 36y \)[/tex]:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \cos(6x) + 36 \sin^2(3x) \][/tex]
6. Simplify the expression further:
Using the identity [tex]\( \sin^2(A) = \frac{1 - \cos(2A)}{2} \)[/tex]:
[tex]\[ \sin^2(3x) = \frac{1 - \cos(6x)}{2} \][/tex]
Therefore:
[tex]\[ 36 \sin^2(3x) = 36 \cdot \frac{1 - \cos(6x)}{2} = 18 - 18 \cos(6x) \][/tex]
7. Combine the terms:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \cos(6x) + 18 - 18 \cos(6x) = 18 \][/tex]
Thus, we have shown:
[tex]\[ \frac{d^2 y}{dx^2} + 36y = 18 \][/tex]
### Part 2: Proving [tex]\( \frac{d^2 z}{dx^2} + 36z = 18 \)[/tex] for [tex]\( z = \cos^2(3x) \)[/tex]
1. Define [tex]\( z \)[/tex]:
[tex]\[ z = \cos^2(3x) \][/tex]
2. Find the first derivative of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex]:
Using the chain rule:
[tex]\[ \frac{dz}{dx} = \frac{d}{dx} [\cos^2(3x)] = 2 \cos(3x) \cdot (-\sin(3x)) \cdot 3 = -6 \sin(3x) \cos(3x) \][/tex]
3. Simplify using a trigonometric identity:
Recall again the double-angle identity:
[tex]\[ -6 \sin(3x) \cos(3x) = -3 \sin(6x) \][/tex]
So:
[tex]\[ \frac{dz}{dx} = -3 \sin(6x) \][/tex]
4. Find the second derivative of [tex]\( z \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d^2 z}{dx^2} = \frac{d}{dx} \left( -3 \sin(6x) \right) = -3 \cdot 6 \cos(6x) = -18 \cos(6x) \][/tex]
5. Construct and simplify the expression [tex]\( \frac{d^2 z}{dx^2} + 36z \)[/tex]:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = -18 \cos(6x) + 36 \cos^2(3x) \][/tex]
6. Simplify the expression further:
Using the identity [tex]\( \cos^2(A) = \frac{1 + \cos(2A)}{2} \)[/tex]:
[tex]\[ \cos^2(3x) = \frac{1 + \cos(6x)}{2} \][/tex]
Therefore:
[tex]\[ 36 \cos^2(3x) = 36 \cdot \frac{1 + \cos(6x)}{2} = 18 + 18 \cos(6x) \][/tex]
7. Combine the terms:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = -18 \cos(6x) + 18 + 18 \cos(6x) = 18 \][/tex]
Thus, we have shown:
[tex]\[ \frac{d^2 z}{dx^2} + 36z = 18 \][/tex]
Hence, both differential equations are satisfactorily proven.
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