Question

verify that the given function y is a solution of the differential equation that follows it. assume that c is an arbitrary constant. y(t)=c_1 sin 9t + c_2 cos 9t; y(t)+81y(t)=0. start by substituting y(t)=c_1 sin 9t + c_2 cos 9t into the second term of y(t)+81y(t)=0. y(t)+□ = 0

Explanation:

Step1: Find the first - derivative of \(y(t)\)

Using the chain - rule, if \(y(t)=C_1\sin(9t)+C_2\cos(9t)\), then \(y^{\prime}(t)=9C_1\cos(9t)-9C_2\sin(9t)\).

Step2: Find the second - derivative of \(y(t)\)

Differentiating \(y^{\prime}(t)\) again using the chain - rule, \(y^{\prime\prime}(t)=- 81C_1\sin(9t)-81C_2\cos(9t)\).

Step3: Substitute \(y(t)\) and \(y^{\prime\prime}(t)\) into the differential equation

Substitute \(y(t)=C_1\sin(9t)+C_2\cos(9t)\) and \(y^{\prime\prime}(t)=-81C_1\sin(9t)-81C_2\cos(9t)\) into \(y^{\prime\prime}(t)+81y(t)=0\).

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Answer:

Since substituting \(y(t)=C_1\sin(9t)+C_2\cos(9t)\) into \(y^{\prime\prime}(t)+81y(t)\) results in \(0\), \(y(t)\) is a solution of the differential equation \(y^{\prime\prime}(t)+81y(t)=0\).