Question

In the previous lecture this method was explained. Recall that an ODE of the type dy/dr+py be rewritten as may 讐-劘塭 dr with ydl/dx- Ipy from where /(x) can be derived The complete solution of this ODE then is a sum of two terms: a term y. which is a solution of the ODE rewritten as d(ly)/dx- lq and a term y2, which follows from solving the homogeneous equation (the ODE with q-0 Task Solve two differential equations and determine the solution y-yi +N2 by the method of İntegrating function. In both cases use y(0)-1 as boundary condition to fix the integration constant. You may use the recurrence formula for n-1 and n -2 to solve a) and b), resp. Derive this formula explicitly, first forn and then n-2

Answer 1

Z dy/dx = 2x - y First order linear ordinary Differential equation y'(x) + p(x) y = q(x) The general solution is y(x) = integral e^integral p(x) dx q(x) dx + c/e^integral p(x) dx substitute dy/dx with y' 2y' = 2x - y Rewrite in the form of first order linear ODE y' = x - 1/2 y y' + 1/2 y = x y'(x) + p(x) y = q(x) p(x) = 1/2, q(x) = x Find the Integrating factor m(x) = e^integral p(x) dx mu(x) = e^integral 1/2 dx = e^x/2 \r\n put the equation in the form (mu(x) y)' = mu(x) q(x) = e^x/2 x = xe^x/2 y(x) = integral xe^x/2 dx + c/e^x/2 integral xe^x/2 dx = > Apply u-substitution u = x/2 du/dx = 1/2 du = 1/2 dx dx = 2 du = > integral 4 e^4 u du Take the constant out: integral a f(x) dx = a integral f(x) dx