a) Find a function y=f(x) that satisfies the differential equation dy/dx = fifth derivative.

This is one of the questions in my practice test, I tried the basic equation of the trig function such as f(x)= sin(x) or f(x)=-sin(x) however in the fourth derivative they reach the same f(x) and the answer at the book says it is these and the cosine functions which will give the fifth derivative equal to the original equation. What is wrong with my working out?

y = sinx

y ' = cosx or dy/dx = cosx
y '' = -sinx
y ''' = -cosx
y '''' = sinx
y ''''' = cosx

so for y = sinx
the first derivative or dy/dx = cosx
and the fifth derivataive = cosx

so f(x) = sinx

To solve the differential equation dy/dx = fifth derivative, we need to find a function whose fifth derivative is equal to itself.

You started with the basic trigonometric functions f(x) = sin(x) and f(x) = -sin(x). However, you noticed that after taking the fourth derivative, both functions become the same. This is true, because the fourth derivative of sin(x) and -sin(x) both result in -sin(x). But keep in mind that we're looking for a function whose fifth derivative is equal to itself.

Let's analyze further. The fifth derivative of sin(x) can be found by taking the derivative four more times, as follows:

1st derivative: f'(x) = cos(x)
2nd derivative: f''(x) = -sin(x)
3rd derivative: f'''(x) = -cos(x)
4th derivative: f''''(x) = sin(x)

Notice that the fourth derivative of sin(x) is not equal to itself, which means sin(x) does not satisfy the given differential equation.

Similarly, you can calculate the fifth derivative of -sin(x) by taking the derivative four more times:

1st derivative: f'(x) = -cos(x)
2nd derivative: f''(x) = sin(x)
3rd derivative: f'''(x) = cos(x)
4th derivative: f''''(x) = -sin(x)

Again, the fourth derivative here (-sin(x)) is not equal to the original function, so -sin(x) does not satisfy the differential equation.

The correct functions that satisfy the differential equation dy/dx = fifth derivative are the trigonometric functions f(x) = A*sin(x) and f(x) = A*cos(x), where A is a constant.

Taking the fifth derivative of f(x) = A*sin(x), we get:

1st derivative: f'(x) = A*cos(x)
2nd derivative: f''(x) = -A*sin(x)
3rd derivative: f'''(x) = -A*cos(x)
4th derivative: f''''(x) = A*sin(x)
5th derivative: f'''''(x) = A*cos(x)

Notice that the fifth derivative (A*cos(x)) is equal to the original function (A*sin(x)). Therefore, f(x) = A*sin(x) satisfies the given differential equation.

Similarly, you can verify that f(x) = A*cos(x) also satisfies the differential equation.

In conclusion, sin(x) and -sin(x) are not the solutions to the differential equation dy/dx = fifth derivative. The correct solutions are f(x) = A*sin(x) and f(x) = A*cos(x), where A is a constant.