If $y=2 x e^{-3 x}$, show that $\frac{d 2 y}{d x^2}+\frac{6 d y}{d x}+9 y=0$
Answer (2)
Find the first derivative d x d y of y = 2 x e − 3 x using the product rule: d x d y = 2 e − 3 x − 6 x e − 3 x .
Find the second derivative d x 2 d 2 y by differentiating d x d y : d x 2 d 2 y = − 12 e − 3 x + 18 x e − 3 x .
Substitute y , d x d y , and d x 2 d 2 y into the equation d x 2 d 2 y + 6 d x d y + 9 y .
Simplify the expression to show that it equals 0: 0 .
Explanation
Problem Analysis We are given the function y = 2 x e − 3 x and asked to show that d x 2 d 2 y + 6 d x d y + 9 y = 0 . This involves finding the first and second derivatives of y with respect to x , and then substituting them into the given equation to verify that it equals zero.
Finding the First Derivative First, we find the first derivative d x d y using the product rule. The product rule states that if y = uv , then d x d y = u ′ v + u v ′ . In our case, u = 2 x and v = e − 3 x . Thus, u ′ = 2 and v ′ = − 3 e − 3 x . Applying the product rule, we get: d x d y = ( 2 ) ( e − 3 x ) + ( 2 x ) ( − 3 e − 3 x ) = 2 e − 3 x − 6 x e − 3 x
Finding the Second Derivative Next, we find the second derivative d x 2 d 2 y by differentiating d x d y with respect to x . We have d x d y = 2 e − 3 x − 6 x e − 3 x . We need to differentiate each term separately. The derivative of 2 e − 3 x is − 6 e − 3 x . For the term − 6 x e − 3 x , we use the product rule again with u = − 6 x and v = e − 3 x . Thus, u ′ = − 6 and v ′ = − 3 e − 3 x . Applying the product rule, we get: d x d ( − 6 x e − 3 x ) = ( − 6 ) ( e − 3 x ) + ( − 6 x ) ( − 3 e − 3 x ) = − 6 e − 3 x + 18 x e − 3 x Therefore, d x 2 d 2 y = − 6 e − 3 x − 6 e − 3 x + 18 x e − 3 x = − 12 e − 3 x + 18 x e − 3 x
Substitution Now, we substitute y , d x d y , and d x 2 d 2 y into the left-hand side of the equation d x 2 d 2 y + 6 d x d y + 9 y :
( − 12 e − 3 x + 18 x e − 3 x ) + 6 ( 2 e − 3 x − 6 x e − 3 x ) + 9 ( 2 x e − 3 x )
Verification Finally, we simplify the expression: − 12 e − 3 x + 18 x e − 3 x + 12 e − 3 x − 36 x e − 3 x + 18 x e − 3 x = ( − 12 + 12 ) e − 3 x + ( 18 − 36 + 18 ) x e − 3 x = 0 e − 3 x + 0 x e − 3 x = 0 Thus, we have shown that d x 2 d 2 y + 6 d x d y + 9 y = 0 .
Conclusion Therefore, we have verified that for y = 2 x e − 3 x , the equation d x 2 d 2 y + 6 d x d y + 9 y = 0 holds true.
Examples
In mechanical engineering, understanding differential equations like the one in this problem is crucial for analyzing damped harmonic motion. For instance, consider a spring-mass system with damping, where the mass's displacement y ( t ) follows a similar differential equation. By solving such equations, engineers can predict and control the system's behavior, ensuring stability and optimal performance in various applications, such as designing suspension systems for vehicles or vibration isolation for sensitive equipment.
By calculating the first and second derivatives of y = 2 x e − 3 x and substituting them into the equation d x 2 d 2 y + 6 d x d y + 9 y = 0 , we demonstrate that this equation is satisfied and equals zero.
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