Recent content by Davidian

I think I understood my mistake. The solutions ##\ u_1(x)## and ##\ u_2(x)## can be obtained by using the formula I posted only if the coefficients of the second order term is equal to 1. In my example, I need to divide everything by a factor 2. Thanks for your help! Davide

Thank you for your reply. The formula is the generic one for the method $$ u_1(x)=-\int\frac{y_2(x)g(x)dx}{W(x)}+c_1, $$ while for $$ u_2(x)=\int\frac{y_1(x)g(x)dx}{W(x)}+c_2, $$ where ##\ g(x)## is the non homogeneous term. It is the same as the solution of the two conditions you posed, apart...

Thanks for the reply: here are the calculations. The characteristic polynomial is given by $$ \lambda^2+\lambda-1=0, $$ whose roots are ##\lambda=-1## and ##\lambda=-1/2##. The solution to the homogeneous problem is $$ y(x)=c_1e^{-x}+c_2e^{x/2}. $$ and the Wronskian is $$...

TL;DR Summary: Variation of parameter VS Undetermined Coefficients Hi all, Suppose we want to solve the following ODE 2y''+y'-y=x+7 with two different methods: undetermined coefficients and variation of parameters. The solutions to the homogeneous problem are given by y_1(x)=exp(-x) and...