I Explanation of all "its linearly independent derivatives"

[Hall]

I'm studying Differential Equations from Tenenbaum's, and currently going through non-homogeneous second order linear differential equations with constant coefficients. Method of Undetermined Coefficients is the concerned topic here. I will put forth my doubt through an example. Let's say we are asked to solve
$$
\begin{align*}
y'' + 4y = x \sin 2x\\
\textrm{complexifying the equation} \\
\tilde{y''} + 4 \tilde{y} = x e^{(2x) i} \\
\end{align*}
$$
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,

Case 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.

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 its 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**?

 

[Mark44]

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.

Your original ODE is second order, so you don't need to go beyond f''(x). The functions f, f', and f'' are linearly independent.

Also, what do you mean by "differentiating our candy"? Is that a typo?

 

[WWGD]

Your original ODE is second order, so you don't need to go beyond f''(x). The functions f, f', and f'' are linearly independent.

Also, what do you mean by "differentiating our candy"? Is that a typo?

He learnt Pdes in the street, Mark44, unlike us ;).

 

[chwala]

I'm studying Differential Equations from Tenenbaum's, and currently going through non-homogeneous second order linear differential equations with constant coefficients. Method of Undetermined Coefficients is the concerned topic here. I will put forth my doubt through an example. Let's say we are asked to solve
$$
\begin{align*}
y'' + 4y = x \sin 2x\\
\textrm{complexifying the equation} \\
\tilde{y''} + 4 \tilde{y} = x e^{(2x) i} \\
\end{align*}
$$
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 its 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**?

Kindly your insight on this;

I think for the homogenous part we shall have;

$y_c (x)= A\cos 2x +B\sin 2x$

then for the inhomogenous part; we can let
$y_p(x)= C[x \sin 2x]$
$y^{'}_p(x)= C[ \sin 2x+2x\cos 2x]$
$y^{''}_p(x)= C[2\cos2x+2\cos2x-4\sin2x]$ ...

$C[2\cos2x+2\cos2x-4\sin2x]+4C[ \sin 2x+2x\cos 2x]=[x \sin 2x]$
$C[4\cos2x-4\sin2x]+4C[ \sin 2x+2x\cos 2x]=[x \sin 2x]$
...

then solving can take normal course with given two boundary conditions...