Understanding the Function and its Derivatives

The problem asks us to find a relationship between the function \(y\) and its second derivative with respect to \(x\). The given function is:

\(y = a \cos 2x + b \sin 2x\)

Here, \(a\) and \(b\) are constants. To find the relationship, we need to calculate the first and second derivatives of \(y\) with respect to \(x\).

Calculating the First Derivative of y

We need to find \(\frac{dy}{dx}\). We will differentiate \(y\) term by term with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}(a \cos 2x + b \sin 2x)\)

Using the linearity of differentiation, we get:

\(\frac{dy}{dx} = a \frac{d}{dx}(\cos 2x) + b \frac{d}{dx}(\sin 2x)\)

Now, we apply the chain rule for differentiating trigonometric functions. Recall that \(\frac{d}{dx}(\cos(kx)) = -k \sin(kx)\) and \(\frac{d}{dx}(\sin(kx)) = k \cos(kx)\).

For \( \frac{d}{dx}(\cos 2x) \), \(k=2\), so the derivative is \(-2 \sin 2x\).

For \( \frac{d}{dx}(\sin 2x) \), \(k=2\), so the derivative is \(2 \cos 2x\).

Substituting these results back into the expression for \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = a (-2 \sin 2x) + b (2 \cos 2x)\)

\(\frac{dy}{dx} = -2a \sin 2x + 2b \cos 2x\)

Calculating the Second Derivative of y

Next, we find the second derivative, \(\frac{d^2y}{dx^2}\), by differentiating the first derivative \(\frac{dy}{dx}\) with respect to \(x\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-2a \sin 2x + 2b \cos 2x)\)

Again, using the linearity of differentiation:

\(\frac{d^2y}{dx^2} = -2a \frac{d}{dx}(\sin 2x) + 2b \frac{d}{dx}(\cos 2x)\)

Applying the chain rule again, as done for the first derivative:

For \( \frac{d}{dx}(\sin 2x) \), \(k=2\), so the derivative is \(2 \cos 2x\).

For \( \frac{d}{dx}(\cos 2x) \), \(k=2\), so the derivative is \(-2 \sin 2x\).

Substituting these results into the expression for \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -2a (2 \cos 2x) + 2b (-2 \sin 2x)\)

\(\frac{d^2y}{dx^2} = -4a \cos 2x - 4b \sin 2x\)

Finding the Differential Equation Relationship

We have found the second derivative:

\(\frac{d^2y}{dx^2} = -4a \cos 2x - 4b \sin 2x\)

We can factor out \(-4\) from the expression:

\(\frac{d^2y}{dx^2} = -4 (a \cos 2x + b \sin 2x)\)

Recall the original function \(y\):

\(y = a \cos 2x + b \sin 2x\)

Notice that the expression in the parenthesis is exactly \(y\). Therefore, we can substitute \(y\) into the equation for the second derivative:

\(\frac{d^2y}{dx^2} = -4y\)

To express this as a standard differential equation, we can move the \(-4y\) term to the left side of the equation:

\(\frac{d^2y}{dx^2} + 4y = 0\)

This equation relates the function \(y\) to its second derivative \(\frac{d^2y}{dx^2}\) and is the differential equation satisfied by the given function.

Comparing with Differential Equation Options

Let's compare our derived differential equation with the given options:

Option 1: \(\frac{{{d^2}y}}{{d{x^2}}} + y = 0\) (Does not match)

Option 2: \(\frac{{{d^2}y}}{{d{x^2}}} + 2y = 0\) (Does not match)

Option 3: \(\frac{{{d^2}y}}{{d{x^2}}} - 4y = 0\) (Does not match)

Option 4: \(\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\) (Matches our result)

The differential equation satisfied by \(y = a \cos 2x + b \sin 2x\) is \(\frac{{{d^2}y}}{{d{x^2}}} + 4y = 0\).

Revision Table: Function and Derivatives Summary

This table summarizes the function and its derivatives calculated above.

Additional Information: Second-Order Linear Homogeneous Differential Equations

The differential equation we found, \(\frac{d^2y}{dx^2} + 4y = 0\), is a specific type of differential equation. It is a second-order (because of the second derivative), linear (y and its derivatives appear with power 1), homogeneous (the right side is 0) differential equation with constant coefficients (the coefficients of \(\frac{d^2y}{dx^2}\) and \(y\) are constants, 1 and 4 respectively).

Such equations often have solutions involving exponential or trigonometric functions. The general form of a second-order linear homogeneous differential equation with constant coefficients is:

\(A \frac{d^2y}{dx^2} + B \frac{dy}{dx} + C y = 0\)

where \(A\), \(B\), and \(C\) are constants. To find the general solution, we usually solve the characteristic equation \(Am^2 + Bm + C = 0\). The roots of this equation determine the form of the solution \(y(x)\).

In our case, \(\frac{d^2y}{dx^2} + 4y = 0\), we have \(A=1\), \(B=0\), and \(C=4\). The characteristic equation is \(m^2 + 4 = 0\). The roots are \(m^2 = -4\), so \(m = \pm \sqrt{-4} = \pm 2i\). When the roots are complex conjugate (\(\alpha \pm i\beta\)), the general solution is of the form \(y(x) = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\). Here, \(\alpha = 0\) and \(\beta = 2\). Thus, the general solution is \(y(x) = e^{0x} (c_1 \cos(2x) + c_2 \sin(2x))\), which simplifies to \(y(x) = c_1 \cos 2x + c_2 \sin 2x\).

This confirms that the given function \(y = a \cos 2x + b \sin 2x\) is indeed the general solution to the differential equation \(\frac{d^2y}{dx^2} + 4y = 0\) (with constants \(a\) and \(b\) being equivalent to \(c_1\) and \(c_2\)). This process shows the connection between a function and the differential equation it satisfies.