Question

Verify that the function satisfies the differential equation. $$ \begin{array}{ll} y=3 \cos x+\sin x & y^{\prime \prime}+y=0 \end{array} $$

Short answer

Yes, the function \(y = 3cos(x) + sin(x)\) verifies and satisfies the second order differential equation \(y'' + y = 0\).

Step-by-step solution

Compute the first derivative

The first derivative of the function \(y = 3cos(x) + sin(x)\) with respect to \(x\) is computed using the definition of derivatives. The derivative of \(cos(x)\) is \(-sin(x)\) and that of \(sin(x)\) is \(cos(x)\). So \[y' = -3sin(x) + cos(x).\]

Compute the second derivative

The second derivative of \(y\), denoted as \(y''\), is derived by differentiating \(y'\) with respect to \(x\). Using the rules of derivatives mentioned in the first step, we get \[y'' = -3cos(x) - sin(x).\]

Substitute into the Differential Equation

Now we substitute \(y\) and \(y''\) into the given differential equation, \(y'' + y = 0\), which yields \[(-3cos(x) - sin(x)) + (3cos(x) + sin(x)) = 0.\] Simplifying this leads to \(0 = 0\), which is a valid equality.

Verifying the solution

Since substituting the derived second derivative and the function itself into the provided differential equation produces a valid mathematical equality, the function \(y = 3cos(x) + sin(x)\) is a solution to the second-order differential equation \(y'' + y = 0\).