Question

Given a second order linear homogeneous differential equation а2(х)у" + а (х)У + аo(х)у — 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions yı, V2. But there are times when only one function, call it y, is available and we would like to find a second linearly independent solution. We can find y2 using the method of reduction of order. First, under the necessary assumption the a2(x) F 0 we rewrite the equation as aт (х) у" + p(x)у + q(x)у —D 0 p(x) — аo(x) a2(x) a2(x)9x) Then the method of reduction of order gives a second linearly independent solution as e-/Px)dx - dx y(x) / y2(x) Cyu Cyi (x) where C is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose example, if we obtain y2 C3e* then we can choose C = 1/3 so that y ex so that all the constants in front reduce to 1. For Given the problem У - 2у + 5у %3D0 and a solution yi e sin(2x) Applying the reduction of order method we obtain the following y(x) Cyu= and e px)dx р(x) %3D So we have e-pix)dx dx (x) / dx = Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at У2 (х) — Суи %3D So the general solution to y - 2y + 4y = 0 can be written as = C1y C2y2 c +C2

Answer 1

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