3
$\begingroup$

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?

|cite|improve this question|||||
$\endgroup$
  • $\begingroup$ 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 '15 at 2:09
3
$\begingroup$

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

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

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ 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 '15 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 '15 at 8:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.