# Explanation of all "its linearly independent derivatives"

• I
• Hall
In summary, the conversation discusses studying differential equations and the method of undetermined coefficients for solving non-homogeneous second order linear differential equations with constant coefficients. An example is used where the equation is complexified and the associated homogeneous differential equation is solved. The topic of linearly independent derivatives is brought up and an example is shown to demonstrate their use in finding a particular solution. However, it is noted that this method may not always be the best alternative for solving the differential equation.

#### 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**?

Hall said:
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?

chwala
Mark44 said:
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
Hall said:
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**?

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...

Last edited:

## 1. What does it mean for derivatives to be linearly independent?

Linear independence refers to a set of derivatives that cannot be expressed as a linear combination of each other. In other words, the derivatives are unique and do not depend on each other.

## 2. Why is it important to have linearly independent derivatives?

Having linearly independent derivatives allows for a more accurate and precise understanding of a function. It also allows for easier manipulation and calculation of the function.

## 3. How can I determine if a set of derivatives are linearly independent?

To determine if a set of derivatives are linearly independent, you can use the Wronskian determinant. If the determinant is non-zero, then the derivatives are linearly independent.

## 4. Can a set of linearly independent derivatives become linearly dependent?

Yes, it is possible for a set of linearly independent derivatives to become linearly dependent. This can happen if the function changes in a way that causes the derivatives to become dependent on each other.

## 5. How does the concept of linear independence apply to higher order derivatives?

The concept of linear independence also applies to higher order derivatives. Just like with first order derivatives, a set of higher order derivatives is linearly independent if they cannot be expressed as a linear combination of each other.

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