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I wanted to solve $$y''-2y'-3y=4e^{-x}+1$$ for which I got the homogeneous solution $$y_h=C_1e^{-x}+C_2e^{3x}, \quad C_1,C_2 \in \mathbb R.$$

Then by plugging particular solution of the form $$y_p=Ae^{-x}+Bx^2+Cx+D$$ and it's derivatives into my differential equation I get $$Ae^{-x}+2B+2Ae^{-x}-4Bx-2C-3Ae^{-x}-3Bx^2-3Cx-3D=4e^{-x}+1$$ which reduces to $$-3Bx^2-x(4B+3C)+2B-2C-3D=4e^{-x}+1$$ which could never hold true since the $A$ terms vanished. What did I do wrong?

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    $\begingroup$ You already have $e^{-x}$ in the homogeneous solution. Then ... $\endgroup$ – Claude Leibovici Jun 22 '19 at 13:48
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    $\begingroup$ Are you sure about the DE having the third derivative $y'''$? $\endgroup$ – user10354138 Jun 22 '19 at 13:49
  • $\begingroup$ @user10354138 thanks, typo $\endgroup$ – Tesla Jun 22 '19 at 13:59
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Hint: Compute the complementary solution by the ansatz $$y=e^{\lambda x}$$ The solution is given by $$y=C_1e^{-x}+C_2e^{3x}$$. For the particular solution make the ansatz $$y_P=a_1+a_2xe^{-x}$$ A possible solution is given by $$y_p=-e^{-x}x-\frac{1}{3}$$

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  • $\begingroup$ The OP has already found the complementary solution. $\endgroup$ – amd Jun 22 '19 at 19:10
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First off, I presume you mean $y''$ instead of $y'''.$ Next, you must modify your guess because both your homogeneous solution and the right-hand side of your ODE contain $e^{-x}$. Instead, try to guess $A+Bxe^{-x}$ for the particular solution.

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$$y''-2y'-3y=4e^{-x}+1$$ You have going your $y_h$ correctly.

For your $y_p$ you need to consider $$y_p=Axe^{-x}+B$$ and find $A$ and $B$

I found $A=-1$ and $B=-1/3$

$$y_p=-xe^{-x}-1/3$$

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$$y''-2y'-3y=4e^{-x}+1\implies (D^2-2D-3)y=4e^{-x}+1\qquad \text{where}\quad D\equiv \frac{dy}{dx}$$

Another approach for P.I.

P.I.$\quad = \frac{1}{D^2-2D-3}(4e^{-x}+1)$

$=4\frac{1}{D^2-2D-3}e^{-x}+\frac{1}{D^2-2D-3}\cdot 1=4e^{-x}\frac{1}{(D-1)^2-2(D-1)-3}\cdot 1-\frac{1}{3}(1+\frac{2}{3}D-\frac{1}{3}D^2)^{-1}\cdot1$

$=4e^{-x}\frac{1}{D^2-4D}\cdot 1-\frac{1}{3}$

$=-e^{-x}\frac{1}{D}(1-\frac{D}{4})^{-1}\cdot 1 -\frac{1}{3}$

$=-e^{-x}\frac{1}{D} \cdot 1 -\frac{1}{3}$

$=-xe^{-x}-\frac{1}{3}$


Consider a differential equation of the form $f(D)y=X$

If $X=e^{ax}$, then

$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$

$2.$ P.I.$\quad =\frac{1}{f(D)}e^{ax}=e^{ax} \frac{1}{f(D+a)}\cdot 1$, if $f(a)= 0$

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