(1 point) Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + ao(x)y=0 we know that a fund
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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 fund
solutions ₁, 2. 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 y₂ using the method of reduction of order. First, under the necessary assumption the a₂(x) #0 we rewrite the equation as Then the method of reduction of order gives a second linearly independent solution as and a solution y₁ = x¹ Applying the reduction of order method to this problem we obtain the following So we have S y" + p(x)y' +q(x)y=0_p(x) = p(x) = Sp(z)dz 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 Y2 = C'3e²¹ then we can choose C = 1/3 so that y₂ = ²¹. Given the problem So the general solution to 9y" - 4y + 4y = 0 can be written as -dx = Y₂(x) = Cy₁u = Cy₁(x) [ª y} (x) =X^8 S a₁(x) a₂(x) › q(x) e-Sp(z)dx y²(x) x²y" + 4xy' - 28y = 0 and e Sp(x)dz y = C₁Y₁ + C2Y2 = C1 dx = dx y} (x) 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 = ao(x) a₂(x)' +c₂
(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