(1 point) Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + a₁(x)y=0 we know that a fund

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1 Point Given A Second Order Linear Homogeneous Differential Equation A X Y A X Y A X Y 0 We Know That A Fund 1 (69.46 KiB) Viewed 43 times

(1 point) Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y' + a₁(x)y=0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions ₁1, 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 = e²² sin(5x) Applying the reduction of order method we obtain the following So we have p(x) = y" + p(x)y' + q(x)y= 0 p(x) J So the general solution to y" - 5y + 4y = 0 can be written as Y₂(x) = Cy₁u = Cy₁(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 Y2 = C'3e²z then we can choose C = 1/3 so that y₂ = ²¹ Given the problem y(x) = s a₁(x) a₂(x) 01/² Y = C₁Y1 + C₂2Y2 = C1 e-Sp(z)dz y} (x) and e-SP(z)dz y" - 4y + 29y = 0 -Sp(z)dz - 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 = q(x) = dx = -dx +6₂₂ ao(x) a₂(x)'