If y= a sinh omega theta + b cosh omega theta, where a, b, omega are constant, prove d²y/d theta ² = omega ² y
Understand the Problem
The question requires proving the relationship between the second derivative of a function y with respect to theta and the original function y, given that y is defined as a linear combination of hyperbolic sine and cosine functions. We need to perform differentiation twice and then verify if the resulting expression equals omega squared times the original function.
Answer
$\frac{d^2y}{d\theta^2} = \omega^2 y$
Answer for screen readers
$\frac{d^2y}{d\theta^2} = \omega^2 y$
Steps to Solve
- Write down the given function
We are given the function $y = A\sinh(\omega \theta) + B\cosh(\omega \theta)$.
- Differentiate $y$ with respect to $\theta$ to find $\frac{dy}{d\theta}$
Using the chain rule and the derivatives of hyperbolic functions, we have: $\frac{d}{d\theta} \sinh(\omega \theta) = \omega \cosh(\omega \theta)$ and $\frac{d}{d\theta} \cosh(\omega \theta) = \omega \sinh(\omega \theta)$. Therefore, $$ \frac{dy}{d\theta} = A\omega\cosh(\omega \theta) + B\omega\sinh(\omega \theta) $$
- Differentiate $\frac{dy}{d\theta}$ with respect to $\theta$ to find $\frac{d^2y}{d\theta^2}$
Differentiating again, we have: $$ \frac{d^2y}{d\theta^2} = A\omega^2\sinh(\omega \theta) + B\omega^2\cosh(\omega \theta) $$
- Relate $\frac{d^2y}{d\theta^2}$ to $y$
Notice that $\frac{d^2y}{d\theta^2} = \omega^2 [A\sinh(\omega \theta) + B\cosh(\omega \theta)]$. Since $y = A\sinh(\omega \theta) + B\cosh(\omega \theta)$, we can write: $$ \frac{d^2y}{d\theta^2} = \omega^2 y $$
$\frac{d^2y}{d\theta^2} = \omega^2 y$
More Information
The second derivative of $y$ with respect to $\theta$ is indeed equal to $\omega^2$ times $y$, as shown in the derivation. This relationship indicates that the function $y$ satisfies a second-order differential equation, commonly found in physics and engineering contexts, particularly in the analysis of oscillatory systems.
Tips
A common mistake is forgetting the chain rule when differentiating $\sinh(\omega \theta)$ and $\cosh(\omega \theta)$. Another mistake is incorrectly differentiating the hyperbolic functions themselves. It's crucial to remember that the derivative of $\sinh(x)$ is $\cosh(x)$ and the derivative of $\cosh(x)$ is $\sinh(x)$.
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