I've solved multiple differential equations in this practice set, and even a few with variation of parameters, but no matter how many times I restart this problem I can't get it. I must be doing something wrong in my approach:
$$y''-y'=e^x.$$ 1) First, I use the homogeneous differential $y'' - y'= 0$, which gives me the complementary solutions: $$y_c=c_1 +c_2e^x.$$ 2) Next, I need to determine the solution using variation of parameters of form: $$y_p = u_1 + u_2 e^x$$ $$y'_p = u'_1 + u'_2e^x + u_2e^x$$ I set $u'_1 + u'_2e^x = 0$; thus, $$ y''_p = u'_2e^x+u_2e^x,$$ and substitute in to the original equation, $y'' - y' = e^x$: $$ u'_2e^x+u_2e^x - u_2e^x = e^x$$ $$u'_2e^x = e^x$$ $$u'_2 = 1$$ And I can solve for the other expressions using substitution ($u'_2e^x = -u'_1$) and integration: $$ u_2 = x, u'_1 = -e^x, u_1 = -e^x$$ And using $y_p = u_1 + u_2e^x = -e^x +xe^x$ and my value for $y_c$: $$y = y_c + y_p = c_1 +c_2e^x - e^x + xe^x$$ This is definitely not the correct solution ($y = c_1 + c_2e^x + xe^x$). What did I do wrong?