Method of Undetermined Coefficients using X's on left hand sideI have a couple questions on this question.The question is asking me to find the general solution tox″+6x′+9x=cos⁡(2t)+sin⁡(2t).Solving for the general solution, I gotYc=C1e−3x+C2xe−3x.I was wondering if the fact that the left hand side uses x's instead of y's matters? For almost every question, it uses y's instead of x's.Also, for Yp. I got Yp=Acos⁡(2t)+Bsin⁡(2t).I ultimately got the answer to beC1e−3x+C2xe−3x−(7cos⁡(2t)169)+(17cos⁡(2t)169),which I do not feel to be correct.

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Method of Undetermined Coefficients using X&#039;s on left hand sideI have a couple questions on this question.The question is asking me to find the general solution tox″+6x′+9x=cos⁡(2t)+sin⁡(2t).Solving for the general solution, I gotYc=C1e−3x+C2xe−3x.I was wondering if the fact that the left hand side uses x&#039;s instead of y&#039;s matters? For almost every question, it uses y&#039;s instead of x&#039;s.Also, for Yp. I got Yp=Acos⁡(2t)+Bsin⁡(2t).I ultimately got the answer to beC1e−3x+C2xe−3x−(7cos⁡(2t)169)+(17cos⁡(2t)169),which I do not feel to be correct.

prangijahnot · Accepted Answer

Step 1In this problem, x is the dependant variable, and t is the independant variable. It looks like for your homogenous solution, you solved the homogenous ODEyc″(x)+6yc′(x)+9yc(x)=0With the resultyc(x)=C1e−3x+C2xe−3xStep 2All the work you did in solving this is totally valid, but you renamed both variables for the ODE. Using the original variables from the problem, we get,xc″(t)+6xc′(t)+9xc(t)=0⇒xc(t)=C1e−3t+C2te−3tYour yp(t) is calculated correctly, but again should actually be named xp(t).You can then get the final answer by adding xc(t) and xp(t).

ze2m1ingkdvu · Accepted Answer

Step 1The complementary solution is xc(t)=C1e−3t+C2te−3t. Note: x is the dependent variable and t is the independent variable in this case.Moreover, the complementary solution is the solution to the following second-order homogeneous differential equation:xc″+6xc′+9xc=0.Step 2Finally, in order to get the particular solution, one must guess it to be:yp(t)=Acos⁡(2t)+Bsin⁡(2t)where A and B are to be found using the Method of Undetermined Coefficients.After substitution and some algebraic manipulation, the equation becomes:(5A+12B)cos⁡(2t)+(−12A+5B)sin⁡(2t)=cos⁡(2t)+sin⁡(2t)This implies one gets the following system of two equations with two unknowns via Method of Undetermined Coefficients:1. 5A+12B=12. −12A+5B=1.Using Cramer’s rule to solve the system, the solution to the system is A=−7169 and B=17169.Step 3This implies the particular solution isxp(t)=−7169cos⁡(2t)+17169sin⁡(2t).Hence, the general solution to the non-homogeneous second-order differential equation is:X(t)=xc(t)+xp(t)=C1e−3t+C2te−3t−7169cos⁡(2t)+17169sin⁡(2t)where C1 and C2 are constants.

Method of Undetermined Coefficients using X's on left

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