Verify whether the given equation(y=c1-cos(x+c2)) is a solution to this equation[(d^2y/dx^2)^2 + (dy/dx)^2 - 1 = 0].
I know that sometimes finding the solution is tricky, but surely just taking the derivatives is easy, no?
y = c1-cos(x+c2)
y' = -sin(x+c2)
y" = -cos(x+c2)
(y")^2 + (y')^2 - 1 = 0
sin^2+cos^2 = 1, right?
To verify whether the given equation y = c1 - cos(x + c2) is a solution to the equation [(d^2y/dx^2)^2 + (dy/dx)^2 - 1 = 0], we need to substitute the given equation into the differential equation and check if it satisfies the equation.
Let's start by finding the first and second derivatives of y with respect to x:
dy/dx = 0 - sin(x + c2) = -sin(x + c2)
(d^2y/dx^2) = -cos(x + c2)
Now, substitute these derivatives back into the differential equation:
[(-cos(x + c2))^2 + (-sin(x + c2))^2 - 1] = 0
Simplifying further:
[cos^2(x + c2) + sin^2(x + c2) - 1] = 0
Using the trigonometric identity cos^2θ + sin^2θ = 1, we have:
[1 - 1] = 0
0 = 0
Since 0 = 0, we can conclude that the given equation y = c1 - cos(x + c2) is indeed a solution to the differential equation [(d^2y/dx^2)^2 + (dy/dx)^2 - 1 = 0].