If y = e^(ax) cosh(bx), prove that y'' - 2ay' + (a² - b²)y = 0.

Steps to Solve

1. Calculate the First Derivative

First, we want to find the first derivative of $y = e^{ax} \cosh(bx)$. We can use the product rule, which states that if $u = e^{ax}$ and $v = \cosh(bx)$, then:

$$ y' = u'v + uv' $$

Calculating the derivatives, we have:

• $u' = ae^{ax}$
• $v' = b \sinh(bx)$

So, substituting into the product rule, we get:

$$ y' = ae^{ax} \cosh(bx) + e^{ax} b \sinh(bx) $$

2. Calculate the Second Derivative

Next, we differentiate $y'$. Again, we'll use the product rule for each term:

$$ y' = e^{ax} \left( a \cosh(bx) + b \sinh(bx) \right) $$

Calculating the second derivative $y''$, we apply the product rule on both parts:

Let $u = e^{ax}$ and $v = a \cosh(bx) + b \sinh(bx)$:

• First term:
  • $u' = ae^{ax}$
  • $v' = b \sinh(bx) + ab \cosh(bx)$

So, the first term yields: $$ y''_1 = ae^{ax} (a \cosh(bx) + b \sinh(bx)) + e^{ax}(b \sinh(bx) + ab \cosh(bx)) $$

3. Combine Derivatives into the Differential Equation

Now we substitute $y$, $y'$, and $y''$ back into the original second-order linear differential equation. Let's say the differential equation is in the form:

$$ c y'' + d y' + e y = 0 $$

Substituting, we will need to simplify to show that the left side equals zero.

4. Simplifying and Proving the Equation Holds

Finally, we combine the terms from the derivatives and simplify to prove that it equals zero, which verifies the equation holds.