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(1 point) Given a second order linear homogeneous differential equation a2(x)y" + a₁(x)y' + ao(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions 31, 32. But there are times when only one function, call it y₁, is available and we would like to find a second linearly independent solution. We can fin 3/2 using the method of reduction of order. First, under the necessary assumption the a2 (x) 0 we rewrite the equation as y" + p(x)y' + q(e)y = 0 p(x) a₁(x) a₂(x)' q(x) Then the method of reduction of order gives a second linearly independent solution as Sp(z)dz 12(2) = Cyu = Cy₁ (2) 1² (2) e dz ao (x) a₂(x)' where is an arbitrary constant. We can choose the arbitrary constant to be anything we like. One useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y/2 C3e² then we can choose C= 1/3 so that y/2 Ee e²x
Given the problem and a solution 3₁ = 2 Applying the reduction of order method to this problem we obtain the following 31(z) = So we have p(z) = e-Sp(z)dz y}(z) [E zy" +8zy - 18y=0 -dz = [ and e plaz dz= Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerical value) we arrive at 32(z) = Cy₁u= So the general solution to 25y" -8y + 4y = 0 can be written as y=₁71 +23/2 = C1 +9