Answer Happy

Given a second order linear homogeneous differential equation a₂(x)y" + a₁(x)y + a(x) y = 0 we know that a fundamental set for this ODE consists of a pair linearly independent solutions y₁, 1₂. 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 a₁(x) a₂(x)' ao (x) a₂(x)' Then the method of reduction of order gives a second linearly independent solution as e-/ p(x)dx y} (x) y" + p(x)y + g(x)y= 0 p(x) = 1₁ (x) / ² 32 (x) = Cy₁u = Cy = Cy₁ (x) q(x) = -dx

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 = C3e²x then we can choose C = 1/3 so that y₂ = ²x. Given the problem x²y" - 7xy + 16y=0 and a solution y₁ = x² Applying the reduction of order method to this problem we obtain the following y} (x) = p(x) = and e- / p(x)dx =

So we have [ -/p(x)dx y} (x) =/ -dx = dx = 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 16y" -y + 4y = 0 can be written as y = C₁y₁ + C₂y2 = C₁ +02