1 Point Given A Second Order Linear Homogeneous Differential Equation A2 X Y A Z Y Ao R Y 0 We Know That A Fun 1 (50.88 KiB) Viewed 17 times
1 Point Given A Second Order Linear Homogeneous Differential Equation A2 X Y A Z Y Ao R Y 0 We Know That A Fun 2 (31.54 KiB) Viewed 17 times
(1 point) Given a second order linear homogeneous differential equation a2(x)y" + a(z)y' + ao(r)y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y/1, 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 find 3/2 using the method of reduction of order. First, under the necessary assumption the a2(z) 0 we rewrite the equation as a₁(x) a₂(x)' q(x) Then the method of reduction of order gives a second linearly independent solution as e 32(x) = Cy₁ (2) º Jp(x)dr y} (x) y" + p(x)y' + g(x)y=0 p(x) = -da ao(x) a₂(x)' where C' is an arbitrary constant. We can choose the arbitrary constant to be anything we like. Once useful choice is to choose C so that all the constants in front reduce to 1. For example, if we obtain y2 = C3e² then we can choose C = 1/3 so that y2 = ²
Given the problem and a solution y₁ = = e(22/5) Applying the reduction of order method to this problem we obtain the following y} (x) = So we have p(x) e Jp(z)dr y} (x) -dx = 25y" 20y + 4y= 0 S and e-p(z)dz da = Finally, after making a selection of a value for C as described above (you have to choose some nonzero numerica value) we arrive at Y2(x)= So the general solution to 25y" - 20y + 4y = 0 can be written as y = C1y1 + €232 = C1 +0₂
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