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Find real-valued fundamental solutions, y1 and y2, of the differential equation:
Find real-valued fundamental solutions, y1 and y2, of the differential equation:y" + 4y' + 13y =0
Use the fundamental solutions found above to write y(t), solution of the differential equation above satisfying the initial conditions:
y(0) = 2, y'(0) = 3
Final answer:Thedifferential equationy'' + 4y' + 13y = 0 has fundamental solutions y1(t) = e^(-2t)cos(3t) and y2(t) = e^(-2t)sin(3t). The solution satisfying initial conditions y(0) = 2 and y'(0) = 3 is y(t) = 2*e^(-2t)cos(3t) + e^(-2t)sin(3t).Explanation:The given differential equation is a second-order, homogeneous,constant coefficient differential equation, which can be solved using the characteristic polynomial method to find the roots. The characteristic polynomial equation for the differential equation y'' + 4y' + 13y = 0 is r^2 + 4r + 13 = 0. Solving this equation using the quadratic formula r = [-b +/- sqrt(b^2 - 4ac)]/(2a) gives you r = -2 +/- 3i. This means that the characteristic roots are complex, which results in a solution in a form of a combination of sine and cosine functions. Thus, thefundamental solutionsy1 and y2 are y1(t) = e^(-2t)cos(3t) and y2(t) = e^(-2t)sin(3t).y(t), solution of the differential equation that satisfies the initial conditionsy(0)= 2 andy'(0)= 3 is found by substituting these initial conditions into y(t) = c1*y1(t) + c2*y2(t) and determining the values for the constants c1 and c2. After substitution and solving, you will find c1 = 2 and c2 = 1. Therefore, the solution is y(t) = 2*e^(-2t)cos(3t) + e^(-2t)sin(3t).Learn more about Differential Equations here:brainly.com/question/33814182#SPJ11...