Question

(1 point) The differential equation dx has r4 as a solution. Applying reduction order we set Y2-uz" Then (using the prime notation for the derivatives) So, plugging 32 into the left side of the differential equation, and reducing, we get The reduced form has a common factor of r5 which we can divide out of the equation so that we have ru" + u0 Since this equation does not have any u terms in it we can make the substitution w-u, giving us the first order linear equation ru, + w-0 This equation has integrating factor If we use a as the constant of integration, the solution to this equation is w- Integrating to get u, and using b as our second constant of integration we have u Finally 2 for x>0 and the general solution is

Answer 1

"Given differential equation is x^2 d^2/dx^2 - 7x dy/dx + 16y = 0 - * y_1 (x) = x^4 Let y_2 (x) = ux^4 y'_2 = x^4 . u^1 + 4ux^3 y^""_2 = x^4 . u' + 4ux^3 y""_2 = x^4 u"" + 4x^3 u' + 4u' x^3 + 12u x^2 Therefore y""_2 = x^4 u"" + 8x^3 u' + 12x^2 . u plugging y_2 into the L.H.S of * we have x^2 y""_2 - 7xy'_2 + 16 y_2 = x^2 (x^4 u"" + 8x^3 u' + 12 x^2 u) - 7x (x^4 u' + 4x^3 u) + 16 x^4 u = x^6 u"" + (8x^5 - 7x^5) u' + (12 x^4 - 28x^4 + 16x^4) u = x^6 u"" + x^5 u' = x^5 (xu"" + u') x^2 y""_2 - 7xy'_2 + 16y_2 = x^5 (xu"" + u') Now x^2 + y""_2 - 7xy'_2 + 16y_2 = 0 Rightarrow x^5 (xu"" + u') = 0 Rightarrow xu"" + u' = 0 - 1 Substitute w = u' so that w' = u"""