Answer Happy

(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 y₁, y2. 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 = e(27/5) Applying the reduction of order method to this problem we obtain the following So we have y" + p(x)y' +q(x)y=0 p(x) p(x) = [= So the general solution to 25y" - 20y + 4y = 0 can be written as 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 = C'3e²z then we can choose C = 1/3 so that y₂ 2x Given the problem dx = Y₂(x) = Cy₁(x) /=/ y₁(x) = a₁(x) a₂(x)' S p(x)dz and e-√p(z)dz y = C₁Y1 + C2Y2 C1 y} (x) 25y" 20y + 4y = 0 -dx -Sp(x)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) = q(x) = dx = ao(x) a₂(x)' +6₂