Dulce Cantrell

2022-09-11

Systems of Differential Equations and higher order Differential Equations.

I've seen how one can transform a higher order ordinary differential equation into a system of first-order differential equations, but I haven't been able to find the converse. Is it true that one can transform any system into a higher-order differential equation? If so, is there a general method to do so?

I've seen how one can transform a higher order ordinary differential equation into a system of first-order differential equations, but I haven't been able to find the converse. Is it true that one can transform any system into a higher-order differential equation? If so, is there a general method to do so?

cerfweddrq

Beginner2022-09-12Added 15 answers

Step 1

If I am understanding your question, you just would reverse the process on the last equation from the system.

An ${n}^{th}$ order diﬀerential equation can be converted into an n-dimensional system of ﬁrst order equations.

There are various reasons for doing this, one being that a ﬁrst order system is much easier to solve numerically (using computer software) and most diﬀerential equations you encounter in “real life” (physics, engineering etc) don’t have nice exact solutions.

Step 2

If the equation is of order n and the unknown function is y, then set:

$${x}_{1}=y,{x}_{2}={y}^{\prime},\dots ,{x}_{n}={y}^{n-1}.$$

Note (and then note again) that we only go up to the $(n-1{)}^{st}$ derivative in this process. Lets do an example in both directions

If I am understanding your question, you just would reverse the process on the last equation from the system.

An ${n}^{th}$ order diﬀerential equation can be converted into an n-dimensional system of ﬁrst order equations.

There are various reasons for doing this, one being that a ﬁrst order system is much easier to solve numerically (using computer software) and most diﬀerential equations you encounter in “real life” (physics, engineering etc) don’t have nice exact solutions.

Step 2

If the equation is of order n and the unknown function is y, then set:

$${x}_{1}=y,{x}_{2}={y}^{\prime},\dots ,{x}_{n}={y}^{n-1}.$$

Note (and then note again) that we only go up to the $(n-1{)}^{st}$ derivative in this process. Lets do an example in both directions

peckishnz

Beginner2022-09-13Added 2 answers

Step 1

Forward Approach

$$\begin{array}{}\text{(1)}& {y}^{(4)}-3{y}^{\prime}{y}^{\u2033}+\mathrm{sin}(t{y}^{\u2033})-7t{y}^{2}={e}^{t}\end{array}$$

Let: ${x}_{1}=y,{x}_{2}={y}^{\prime},{x}_{3}={y}^{\u2033},{x}_{4}={y}^{\u2034}$ and substitute into (1), yielding:

- ${x}_{1}^{\prime}={y}^{\prime}={x}_{2}$

- ${x}_{2}^{\prime}={y}^{\u2033}={x}_{3}$

- ${x}_{3}^{\prime}={y}^{\u2034}={x}_{4}$

- ${x}_{4}^{\prime}={y}^{(4)}=3y{y}^{\u2033}-\mathrm{sin}(t{y}^{\u2033})+7t{y}^{2}+{e}^{t}=3{x}_{2}{x}_{4}-\mathrm{sin}(t{x}_{3})+t{x}_{1}^{2}+{e}^{t}$

Step 2

Looking at the last equation from the system, we let: $y={x}_{1},{y}^{\prime}={x}_{2},{y}^{\u2033}={x}_{3},{y}^{\u2034}={x}_{4}$ and substitute into the system's last equation above, yielding:

- ${y}^{(4)}=3{y}^{\prime}{y}^{\u2033}-\mathrm{sin}(t{y}^{\u2033})+7t{y}^{2}+{e}^{t}$

Forward Approach

$$\begin{array}{}\text{(1)}& {y}^{(4)}-3{y}^{\prime}{y}^{\u2033}+\mathrm{sin}(t{y}^{\u2033})-7t{y}^{2}={e}^{t}\end{array}$$

Let: ${x}_{1}=y,{x}_{2}={y}^{\prime},{x}_{3}={y}^{\u2033},{x}_{4}={y}^{\u2034}$ and substitute into (1), yielding:

- ${x}_{1}^{\prime}={y}^{\prime}={x}_{2}$

- ${x}_{2}^{\prime}={y}^{\u2033}={x}_{3}$

- ${x}_{3}^{\prime}={y}^{\u2034}={x}_{4}$

- ${x}_{4}^{\prime}={y}^{(4)}=3y{y}^{\u2033}-\mathrm{sin}(t{y}^{\u2033})+7t{y}^{2}+{e}^{t}=3{x}_{2}{x}_{4}-\mathrm{sin}(t{x}_{3})+t{x}_{1}^{2}+{e}^{t}$

Step 2

Looking at the last equation from the system, we let: $y={x}_{1},{y}^{\prime}={x}_{2},{y}^{\u2033}={x}_{3},{y}^{\u2034}={x}_{4}$ and substitute into the system's last equation above, yielding:

- ${y}^{(4)}=3{y}^{\prime}{y}^{\u2033}-\mathrm{sin}(t{y}^{\u2033})+7t{y}^{2}+{e}^{t}$

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