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?

  • $\begingroup$ The method of undetermined coefficients might be easier here. $Y_p=x(Ax+B)e^x$. Take the derivatives and plug in to solve for $A$ and $B$. You can also simply drop the $-e^x$ from your solution since it solves the homogeneous. You can always drop anything that solves the homogeneous equation from your particular solution as it contributes nothing. $\endgroup$
    – jdods
    Mar 5, 2015 at 2:09

1 Answer 1


$$y = y_c + y_p = c_1 +C_2e^x - e^x + xe^x$$ $C_2e^x - e^x=(C_2-1)e^x=c_2e^x$ where $c_2=C_2-1$

So, you did nothing wrong : $$y = c_1 + c_2e^x + xe^x$$

  • $\begingroup$ I'm afraid I don't understand the rationale behind defining $c_2$ as $C_2-1$. Does $C_2$ not equal $c_2$? $c_2$ is a constant and if I'm using the same value in my variation of parameters equation then it wouldn't change. $\endgroup$
    – Alex
    Jan 21, 2015 at 7:50
  • $\begingroup$ The constant is arbitrary. It can be any value. So, you could rempace $c_2$ by $(c_2+3)$ which is also any constant. Or you could replace it by $(c_2-c_3)$ or by $-c_2$ or by $(c_2-1)$, etc. All are equivalent and all formula with the different writtings of constant are correct. Instead of $c_1$ and $c_2$ the other author could have used other notations for the constants, for example $a_1$ and $a_2$, or $C_1$ and $C_2$. $\endgroup$
    – JJacquelin
    Jan 21, 2015 at 8:09

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