- #1

- 351

- 87

$$

\begin{align*}

y'' + 4y = x \sin 2x\\

\textrm{complexifying the equation} \\

\tilde{y''} + 4 \tilde{y} = x e^{(2x) i} \\

\end{align*}

$$

If we see the associated homogeneous differential equation

$$

\begin{align*}

y'' + 4y = 0 \\

\textrm{Characteristic equation} \\

m^2 + 4 = 0 \\

m = \pm 2i \\

y_c = c e^{(2i) x} \\

\end{align*}

$$

Now, Tenenbaum writes,

So, in our case ##u(x) = e^{(2x) i}## and ##k=1##; ##y_p## should be a linear combination of ##x^2 e^{(2i) x}## and itsCase 2.##Q(x)## [the RHS of a non-homogeneous differential equation] contains a term which, ignoring constant coefficients, is ##x^k## times a term ##u(x)## of ##y_c##, where ##k## is zero or a positive integer.In this case a particular solution ##y_p## of [non-homogeneous differential equattion] will be a linear combination of ##x^{k+1} u(x)## and all its linearly independent derivatives.

**linearly independent derivatives.**What are its linearly independent derivatives? A set is called independent if none of the members can be obtained by combining the rest of them linearly. Let's begin differentiating our candy

$$

\begin{align*}

f(x) = x^2 e^{(2x) i } \\

f'(x) = 2x e^{(2x) i } + 2i x^2 e^{(2x) i } \\

f''(x) = 2 e^{(2x) i } + 8i x e^{(2x) i }-4 x^2 e^{(2x) i } \\

f'''(x) = 12i e^{(2x) i } - 24 x e^{(2x) i } - 8i x e^{(2x) i }\\

\end{align*}

$$

Well, it seems to me that ##f'''(x)## can be expressed as linear combination of ##f"(x)## and ##f'(x)## (but I'm not sure. So, according to Tenenbaum, the solution should like

$$

y_p = A x^2 e^{(2x) i } + 2Bx e^{(2x) i } + 2B i x^2 e^{(2x) i } + 2C e^{(2x) i } + 8C i x e^{(2x) i } - 4C x^2 e^{(2x) i }

$$

?

I'm not sure if I'm right, but this task of finding linearly independent derivatives doesn't seem a very good alternative to solving the actual DE. Can you please say something on this **linear independent derivatives**?