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Solve the given differential equation by the undetermined coefficients method:

$$y'' + 6y = −294x^2e^{6x}.$$

For this problem I got the answer as: c $$C_2 \sin (\sqrt{6x}) + C_1\cos(\sqrt{6x}) + e^{6x}(-7x^2 + 4x -17/21)$$

but I still didn't get the right answer for it. I'm Not sure what's going wrong. Thanks.

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    $\begingroup$ Don't use all caps in a title. Caps lock is not, in fact, cruise control for "cool." $\endgroup$ Feb 12, 2015 at 4:35
  • $\begingroup$ @CameronWilliams Comedy gold. $\endgroup$
    – mattos
    Feb 12, 2015 at 4:36
  • $\begingroup$ Notice that your homogeneous solution arguments should be $\sqrt{6}x$ not $\sqrt{6x}$. And I'm guessing it's your particular solution that you made the mistake on. What ansatz did you take? $\endgroup$
    – mattos
    Feb 12, 2015 at 4:40
  • $\begingroup$ I got it. Thanks you guys. I just had sqrt6 in a wrong format. $\endgroup$
    – user214862
    Feb 12, 2015 at 5:56

2 Answers 2

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Check by plugging the solution in the equation. The homogeneous part certainly fits (after fixing the square root typo). Then for the non-homogeneous part,

$$y''+6y=36e^{6x}(-7x^2 + 4x -17/21)+12e^{6x}(-7x^2 + 4x -17/21)(-14x+4)+e^{6x}(-14)+e^{6x}(-7x^2 + 4x -17/21)\\ =-294x^2e^{6x}.$$

Well done.

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For the complementary solution, you need to solve $y"+6y=0$. Use the substitution $y=e^{\lambda x},~\lambda >0$ and some manipulation to get $\lambda=i\sqrt{6}$ or $\lambda=-i\sqrt{6}$ which imply $y_1=C_1e^{i\sqrt{6}x},~y_2=C_2e^{-i\sqrt{6}x}$. Applying Euler's identity and regrouping terms gives us \begin{equation*} y=c_3\cos(\sqrt{6}x)+C_3\sin(\sqrt{6}x). \end{equation*} The particular solution will be of the form \begin{equation*} y_p=A_1e^{6x}+A_2e^{6x}x+A_3e^{6x}x^2. \end{equation*} Differentiate this twice and insert into the differential equation. Simplify and equate the coefficients. You should get the particular solution to be \begin{equation*} y_p=-\frac{17e^{6x}}{21}-7e^{6x}x^2+4e^{6x}x. \end{equation*} Try it again and let me know if you are still having problems.

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