I did a question that was given only 1 out of 2 marks and I did not know which part I went wrong. The question is:

$\frac{d^{2}y} {dx^2} + 4 \frac {dy}{dx} + 4y = 25 \sin(x) + e^{-2x} $

I answered as:

Complimentary function (using auxillary equation): $ k^2 + 4k + 4 =0 $ $ k=-2 $ twice $ Y_{cf} = Ae^{-2x} + Bxe^{-2x}$

Particular Integral: Guess: $Y_p = C_1\sin(x) + C_2 \cos(x) + C_3e^{-2x}$

This is redundant to the complimentary function. $Y_p = C_1\sin(x) + C_2 \cos(x) + C_3x^2e^{-2x}$ $Y_p' = C_1\cos(x) - C_2 \sin(x) -2 C_3x^2e^{-2x} +2C_3xe^{-2x}$ $Y_p ''= -C_1\sin(x) + C_2 \cos(x) + 2C_3e^{-2x} +4C_3x^2e^{-2x} -8 C_3xe^{-2x}$

Substitute to original equation gives: $(3C_1 - 4C_2)\sin(x) + (3C_2+4C_1)\cos(x) +2C_3 e^{-2x} = 25\sin(x) +e^{-2x}$

Then I have $C_1 = -\frac{75}{28}, C_2 = \frac{75}{7}, C_3= \frac{1}{2} $

Thus $Y(x) = Ae^{-2x} +Bxe^{-2x} - \frac{75}{28} \sin(x) + \frac{75}{7} \cos(x) + \frac{1}{2}x^2e^{-2x} $

Did I do something wrong?


Let us generalise a bit: we have a linear differential equation with constant coefficients $$y''+ay'+b=e^{cx}.$$ Suppose that $c$ is not a root of the characteristic polynomial $p(z) = z^2+az+b.$ Then a particular solution writes $$y_{particular}(x) = \frac{e^{cx}}{p(c)}.$$

In your case $p(z) = z^2+4z+4$; we write the right hand side as $\frac{25}{2i}(e^{ix}-e^{-ix}) + e^{-2x}$. Thus, we already know two terms of the particular solution: $$\frac{25}{2i} \frac{e^{ix}}{p(i)}-\frac{25}{2i}\frac{e^{-ix}}{p(-i)}=\frac{25}{2i} \frac{e^{ix}}{3+4i}-\frac{25}{2i}\frac{e^{-ix}}{ 3-4i}$$$$=25 \Im\left(\frac{e^{ix}}{3+4i}\right) =\Im(3e^{ix}-4ie^{ix})=3\sin x-4\cos x.$$

Then again, if $c$ is a root of $p$ with multiplicity $k$, then the particular solution can be guessed in the form $\alpha x^{k+1}e^{cx}$ In your case, indeed, you need to study $x^2e^{-2x}$: after the simplifications, we obtain that, as you already found out, $\alpha=1/2$.

Therefore, the final solution writes $$3\sin x-4\cos x +\frac{x^2e^{-2x}}{2} + Ae^{-2x} + Bxe^{-2x}. $$

You error, I suppose, is in finding the solutions of the linear system $$4c_1+3c_2=0\\3c_1-4c_2=25.$$


Your method is right, but you made a mistake in your final solution. The coefficients for $\sin x$ and $\cos x$ are:

$$ 3C_1 - 4C_2 = 25 $$ $$ 4C_1 + 3C_2 = 0 $$

Solving that should give $C_1 = 3$ and $C_2 = -4$


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