show that any fuction

y(t)=acosx+bsinx ,satisfied the differential equation y^"+wy=0 and also find the value of (a) and (b)

To show that the function y(t) = a*cos(x) + b*sin(x) satisfies the differential equation y'' + wy = 0, we need to differentiate y(t) twice and substitute it into the differential equation. Let's start by differentiating y(t):

y(t) = a*cos(x) + b*sin(x)

First derivative:
y'(t) = -a*sin(x) + b*cos(x) (using the chain rule)

Second derivative:
y''(t) = -a*cos(x) - b*sin(x) (again using the chain rule)

Now, substitute y(t), y'(t), and y''(t) into the differential equation:

y''(t) + wy(t) = 0
[-a*cos(x) - b*sin(x)] + w[a*cos(x) + b*sin(x)] = 0

Now, let's simplify this equation:

(-a + wa)*cos(x) + (-b + wb)*sin(x) = 0

For this equation to hold true for all x, the coefficients of cos(x) and sin(x) must be zero:

-a + wa = 0
-b + wb = 0

By solving these two equations, we can find the values of a and b that satisfy the differential equation.

From the first equation, we have:
-a + wa = 0
a(1 - w) = 0

This equation is satisfied either when a = 0 or when w = 1. However, if w = 1, the differential equation becomes y'' + y = 0, which is not the same equation we started with. Therefore, we conclude that a = 0.

From the second equation, we have:
-b + wb = 0
b(1 - w) = 0

This equation is satisfied either when b = 0 or when w = 1. As mentioned earlier, we disregard the case when w = 1. Therefore, b = 0.

Therefore, the values of a and b that satisfy the differential equation y'' + wy = 0 for the function y(t) = a*cos(x) + b*sin(x) are a = 0 and b = 0.