Given that y= Acos3x + Bsin3x , where A and B are constants, show that d^2y/dx^2 + 9y =0
To show that d^2y/dx^2 + 9y = 0, we'll need to take the second derivative of y with respect to x and then substitute it into the equation.
First, let's find dy/dx. Taking the derivative of y = Acos3x + Bsin3x with respect to x, we have:
dy/dx = -3Asin3x + 3Bcos3x
Next, we'll find d^2y/dx^2. Taking the derivative of dy/dx with respect to x, we have:
d^2y/dx^2 = -9Acos3x - 9Bsin3x
Now, let's substitute d^2y/dx^2 into the equation d^2y/dx^2 + 9y = 0:
(-9Acos3x - 9Bsin3x) + 9(Acos3x + Bsin3x) = 0
Simplifying this equation, we have:
-9Acos3x - 9Bsin3x + 9Acos3x + 9Bsin3x = 0
The terms -9Acos3x and 9Acos3x cancel out, as well as the terms -9Bsin3x and 9Bsin3x. This leaves us with:
0 = 0
This proves that d^2y/dx^2 + 9y = 0 is satisfied for the function y = Acos3x + Bsin3x, where A and B are constants.
y" = -9Acos3x - 9Bsin3x
it follows
note that for any linear combination of sin(ax) and cos(ax)
y" + a^2 y = 0