# Reduction of Order

May 8th, 2015
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Consider a linear, second-order, homogeneous equation in standard form, y 00 + p(x)y 0 + q(x)y 0 = 0: Suppose that one solution y1 is known. Then a second, independent solution y2 is obtained by letting y2 (x) = u(x)y1 (x); u(x) to be determined

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Math(diff. equations) notes #1MATH 1005A - Notes 1Reduction of OrderConsider a linear, second-order, homogeneous equation in standard form, y 00 + p(x)y 0 + q(x)y 0 = 0:Suppose that one solution y1 is known. Then a second, independent solution y2 is obtained by letting y2 (x) = u(x)y1 (x); u(x) to be determined.00000y2 = uy1 ) y2 = u0 y1 + uy1 ; y 00 = u00 y1 + 2u0 y1 + uy1 ;so y2 is a solution if and only if0000[u00 y1 + 2u0 y1 + uy1 ] + p(x)[u0 y1 + uy1 ] + q(x)[uy1 ] = 0;i.e.,0000u[y1 + p(x)y1 + q(x)y1 ] + u00 y1 + 2u0 y1 + p(x)u0 y1 = 0:000Since y1 is a solution, y1 + p(x)y1 + q(x)y1 = 0. Thus, y2 is a solution if and only if 0u00 y1 + u0 [2y1 + p(x)y1 ] = 0;i.e.,u002y 0 + p(x)y1y0= 1= 2 1 p(x):u0y1y1Integration with respect to x then gives0ln ju j = 2 ln jy1 j Zp(x) dx:Taking the exponential of both sides and using the fact that e2 ln jy1 j = eln jy1 j obtainju0 j =2=1, we2y11 R p(x) dx1 Re; or u0 = 2 e p(x) dx:2y1y1Taking the plus sign and integrating once more, we obtainZ1 R p(x) dxu(x) =edx:2y1Since1 R p(x) dxe6= 0;2y1u(x) is not a constant, hence y1 = erx and y2 = u(x)erx are linearly independent. For the equation ay 00 + by 0 + cy = 0 with b2 4ac = 0; y = erx ) ar2 + br + c = 0 ) pbbb b2 4ac=) y1 = erx = e 2a x :r=2a2a2In standard form, the equation isbcby 00 + y 0 + y = 0; with p(x) = :aaaThus, by reduction of order,

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