1 Point Given A Second Order Linear Homogeneous Differential Equation Az R Y A1 X Y Ao X Y 0 We Know That A Fun 1 (43.65 KiB) Viewed 14 times
1 Point Given A Second Order Linear Homogeneous Differential Equation Az R Y A1 X Y Ao X Y 0 We Know That A Fun 2 (32.74 KiB) Viewed 14 times
(1 point) Given a second order linear homogeneous differential equation az (r)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 find 3/2 using the method of reduction of order. First, under the necessary assumption the a2 (a) 0 we rewrite the equation as y" + p(x)y' + q(a)y=0 p(x) q(x) a₂(x)' Then the method of reduction of order gives a second linearly independent solution as e-Jp(z)dz 32 (2) = Cy₁u = Cy₁ (2) 2 -da y} (x) ao (x) a₂(x)' where C 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 e²z C=1/3 so that y/2 =
Given the problem and a solution y/1 e³ sin(2x) Applying the reduction of order method we obtain the following y(x) = So we have = p(x) -Sp(z)dz y} (x) 1= -da y" - 6y + 13y = 0 1 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 numerical value) we arrive at y2(x) = Cy₁u = So the general solution to y" - 2y + 4y = 0 can be written as y = C1y1 + 2Y/2 = C1 +0₂
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