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For the given differential equation, (a) Determine the complementary solution, yc(t)=c1y1(t)+c2y2(t). (b) Use the method of variation of parameters to construct a particular solution. Then form the general solution. $$ y^{\prime \prime}+y=\sec t, \quad-\pi / 2

Short Answer

Expert verified
Question: Determine the complementary solution, specific solution, and general solution for the given second order linear nonhomogeneous differential equation: y+y=sect, with π/2<t<π/2. Answer: The complementary solution is given by yc(t)=c1cost+c2sint. The particular solution is given by yp(t)=cos2ttsint. The general solution is given by y(t)=c1cost+c2sintcos2ttsint.

Step by step solution

01

Identify the homogeneous equation

The given equation is a nonhomogeneous equation of the form y+y=g(t), where g(t)=sect. We first need to find the general solution of the corresponding homogeneous equation, which is given by y+y=0.
02

Solve the homogeneous equation

The auxiliary equation for this homogeneous equation is given by m2+1=0. Solving for m, we have m=±i. So the characteristic roots are m1=i and m2=i. The general solution of the homogeneous equation is given by: yc(t)=c1cost+c2sint where c1 and c2 are arbitrary constants. #b) Use the method of variation of parameters to construct a particular solution#
03

Calculate Wronskian

We need to calculate the Wronskian W(y1,y2) of the solutions for the homogeneous equation, where y1(t)=cost and y2(t)=sint. The Wronskian is given by: W(y1,y2)=|costsintsintcost|=cos2t+sin2t=1
04

Using the method of variation of parameters

According to the method of variation of parameters, the particular solution for the nonhomogeneous equation is given by: yp(t)=y1(t)y2(t)g(t)W(y1,y2)dt+y2(t)y1(t)g(t)W(y1,y2)dt where g(t)=sect. So, we need to compute the following integrals: sintsectdt=sintcostdt=d(cost)=cost+k1 and costsectdt=1dt=t+k2 Now, substituting these integrals in the equation of yp(t): yp(t)=cost(cost+k1)+sint(t+k2) Since k1 and k2 are arbitrary constants, we can drop them from the particular solution, as they will be absorbed into the general solution. Thus, our particular solution is: yp(t)=cos2ttsint
05

Forming the general solution

The general solution is given by the sum of the complementary solution and the particular solution: y(t)=yc(t)+yp(t)=c1cost+c2sintcos2ttsint

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complementary Solution
The complementary solution, often symbolized as yc(t), is the general solution to the corresponding homogeneous differential equation. In the context of solving differential equations, when we have a nonhomogeneous differential equation like y+y=g(t), the first step is to ignore the nonhomogeneous part (the g(t) term) and solve the homogeneous equation, y+y=0.

Upon finding the characteristic equation, in our case m2+1=0, we determine the roots. For complex roots m1=i and m2=i, the solution involves sine and cosine functions. Therefore, the complementary solution for our exercise is yc(t)=c1cos(t)+c2sin(t), where c1 and c2 are arbitrary constants that will later be determined by initial conditions or additional constraints. This solution captures the behavior of the system's free response without the influence of the external force or input represented by g(t).
Method of Variation of Parameters
Moving beyond the complementary solution, we introduce the method of variation of parameters to find a particular solution to the nonhomogeneous differential equation. This technique is useful when the nonhomogeneous term is not easily manageable by other methods, such as undetermined coefficients.

The method of variation of parameters requires the solutions for the homogeneous equation, which we already obtained, and it involves calculating integrals to find a particular solution yp(t) that will satisfy the original nonhomogeneous equation. This method is dependent on determining a function for each term in the complementary solution, by which those terms are varied to accommodate the nonhomogeneity of the equation. The integration process involves the Wronskian and can result in a particular solution that looks quite different from the complementary solution.
Wronskian
The Wronskian is a determinant used in the analysis of differential equations, named after the Polish mathematician Józef Hoene-Wroński. It is a function that helps in verifying the linear independence of solutions to differential equations, which is crucial for the method of variation of parameters.

In our case, to find the particular solution using variation of parameters, we calculate the Wronskian of the two homogeneous solutions y1(t) and y2(t), which are cos(t) and sin(t) respectively. The Wronskian is the determinant of a matrix composed of these functions and their derivatives. As shown in the solution, W(y1,y2) is equal to 1, which confirms the linear independence of y1(t) and y2(t), a necessary condition to proceed with the method of variation of parameters.
Particular Solution
A particular solution yp(t) to a differential equation addresses the nonhomogeneous aspect of the equation, yielding a solution that works for the entire equation, not just its homogeneous part. It represents a specific response to the external input or force g(t) featured in the nonhomogeneous equation.

In the exercise, once we obtained yp(t)=cos2(t)tsin(t), we've effectively found a solution that includes the effect of sec(t), the nonhomogeneous part of our original differential equation. The complete response of the system is then described by the general solution, which combines the complementary solution yc(t) and the particular solution we've found, leading to y(t)=yc(t)+yp(t). This results in a formula that can model the full behavior of the system governed by the differential equation, subject to necessary initial conditions or boundary values.

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