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12. Linear Differential Equations 319 Let i = (h. (0), (0)) = (1, 2) and 12 = (12(0), 7(0)) = (1,3) be the initial conditions. These two vectors are evidently independent. Thus every possible vector of initial conditions is a linear combination of Ti and 15. From this, we see that every solution of the homogeneous DE is of the form h(x) = det her Now let us return to the inhomogeneous problem. A technique called the method of undetermined coefficients works well here. This is just a fancy name for good guesswork. It works for forcing functions that are (sums of) exponentials, polynomi- als, sines, and cosines. We look for a solution of the same type. Here we hypothesize a solution of the form r) = sir: A + 4 48x, where c and d are constants. Plug finto our differential equation: " -5 -6 = (-c sinx-d cosx)-5cc052-d sinx]-6ệc sinx+d cosx) 5c5d] sinx+54 5c] COS X. So we may solve the system of linear equations 5c+5d-1, 5c 5d = 0, to obtain e-d-0.1. This is a particular solution. Now the general solution to the inhomogeneous equation is of the form f(x) = (0,1 sinx -- {).I cosx- aer+be. We compute the initial conditions 1 = f(0) -0.1+a+b. 0=/(0) - 0.1 +2a+3b. Solving this linear system yields a -2.8 and b=-1.9. Thus the solution is f(x) - 0.1 sinx+0.1 cos.x+2.8e24 -1.9 Exercises for Section 12.6 A. Solve y" - 3y - 10y = 801 – 3, and y(0) = 0). B. Consider y + by' + cy=0. Factor the quadraticx-bx-c=(x-7)(x--s). (a) Solve the DE when rands are distinct real rouls. (b) When r = a +ib and s= a -- ib are distinct complex roots, show that ek sin box and cosax are solutions (c) When ris a double real root, show that and we are solutions, c. Observe that X is a solution of y'-x-2y+x-3y = 0. Look for a second solution of the form ) = x(x) HINT: Find a first-order DE for y.