Please do 2 I have supplied the general form of 2.27 2.3 Nonhomogeneous Equationsdifferent frequencies. If the forcing frequency w is close to the natural frequencyThis solution shows that the charge response is a sum of two oscillations ofthen the amplitude is bounded, but it is obviously large because of the factorw-woccurring in the denominator. Thus the system has large oscillationswhen w is close to wo.113Example 2.27In the previous example, what happens if win (2.17) is invalid because of division by zero; thus we have to re-solve thewo? Then the general solutionproblem. The circuit equation isQ"+wQ = sin wot,(2.19)where the circuit is forced at the same frequency as its natural frequency. Thehomogeneous solution is the same as before, but the particular solution nowhas the formQp(t)t(A sin wot + B cos wot),with a factor of t multiplying the terms. Therefore the general solution of (2.19)has the formQ(t) = C₁ coswot + c2 sin wot + t(A sin wot + B cos wot).Without actually determining the constants, we can infer the nature of theresponse. Because of the t factor in the particular solution, the amplitude ofthe oscillatory response (t) grows in time. This is the phenomenon of pureexternal force is the same as the natural frequency of the system.resonance, or mathematical resonance. It occurs when the frequency of theThe previous example is an ideal case and physically unreasonable. Alwhat happens if we include a small damping term in the circuit and still forccircuits have resistance, or dissipation, even though it may be small. We aslit at itn Ō-1-20102030t405060Figure 2.8 Plot of the solution of Q" +0.2Q' +2Q = sin(√2 t) with zeroinitial conditions. The system is driven at a frequency equal to the naturalfrequency, and there is small damping.EXERCISES1. Plot the solution (2.18) for several different values of 3 and w. Includevalues where these two frequencies are close.2. Find the solution in Example 2.27 if the initial conditions are Q(0) =Q'(0) = 0.3. Find the form of the general solution of the equation I" + 161 = cos 4t.

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Please do 2 I have supplied the general form of 2.27 2.3 Nonhomogeneous Equationsdifferent frequencies. If the forcing frequency w is close to the natural frequencyThis solution shows that the charge response is a sum of two oscillations ofthen the amplitude is bounded, but it is obviously large because of the factorw-woccurring in the denominator. Thus the system has large oscillationswhen w is close to wo.113Example 2.27In the previous example, what happens if win (2.17) is invalid because of division by zero; thus we have to re-solve thewo? Then the general solutionproblem. The circuit equation isQ&#34;+wQ = sin wot,(2.19)where the circuit is forced at the same frequency as its natural frequency. Thehomogeneous solution is the same as before, but the particular solution nowhas the formQp(t)t(A sin wot + B cos wot),with a factor of t multiplying the terms. Therefore the general solution of (2.19)has the formQ(t) = C₁ coswot + c2 sin wot + t(A sin wot + B cos wot).Without actually determining the constants, we can infer the nature of theresponse. Because of the t factor in the particular solution, the amplitude ofthe oscillatory response (t) grows in time. This is the phenomenon of pureexternal force is the same as the natural frequency of the system.resonance, or mathematical resonance. It occurs when the frequency of theThe previous example is an ideal case and physically unreasonable. Alwhat happens if we include a small damping term in the circuit and still forccircuits have resistance, or dissipation, even though it may be small. We aslit at itn Ō-1-20102030t405060Figure 2.8 Plot of the solution of Q&#34; +0.2Q' +2Q = sin(√2 t) with zeroinitial conditions. The system is driven at a frequency equal to the naturalfrequency, and there is small damping.EXERCISES1. Plot the solution (2.18) for several different values of 3 and w. Includevalues where these two frequencies are close.2. Find the solution in Example 2.27 if the initial conditions are Q(0) =Q'(0) = 0.3. Find the form of the general solution of the equation I&#34; + 161 = cos 4t.

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