the equation y+y"=0 is the differential equation of?
a.)sine function
b.)sinusoidal
c.)cosine function
d.) a and c
e.) abc
sine or cosine, since
(sin x)" = -sin x
(cos x)" = -cos x
So, (D)
To determine the differential equation represented by the equation y + y" = 0, we need to understand the given equation's form. The equation y + y" = 0 is a second-order linear homogeneous differential equation.
Here's an explanation of the terms:
- Second-order: The highest derivative of y in the equation is of order 2 (y").
- Linear: The equation is linear in terms of y and its derivatives, meaning there are no multiplication or power operations involving y or its derivatives.
- Homogeneous: The equation is homogeneous because all the terms involve y or its derivatives, and there are no constant terms.
To find the type of function that satisfies this differential equation, we can look for a general solution.
Assume a general solution of the form y = e^(rx), where r is a constant to be determined.
Taking the first derivative of y with respect to x, we get:
y' = re^(rx)
Taking the second derivative of y with respect to x, we get:
y" = r^2e^(rx)
Now, substitute these expressions into the differential equation:
y + y" = e^(rx) + r^2e^(rx) = 0
Since e^(rx) is never zero for any value of x, we can divide the whole equation by e^(rx):
1 + r^2 = 0
To solve for r, we can subtract 1 from both sides of the equation:
r^2 = -1
This is a quadratic equation with complex solutions. Taking the square root of both sides, we obtain:
r = ±sqrt(-1) = ±i
As a result, the solutions for r are purely imaginary, representing complex numbers.
Therefore, the general solution of the given differential equation is of the form:
y = c1*cos(x) + c2*sin(x),
where c1 and c2 are arbitrary constants.
From these calculations, we can conclude that the differential equation y + y" = 0 is the differential equation of both the sine function (sin(x)) and the cosine function (cos(x)).
Hence, the correct answer is option (d): a and c.