- #1

epkid08

- 264

- 1

Edit: It has to be homogeneous.

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- Thread starter epkid08
- Start date

In summary, to find the simplest differential equation with a given set of particular solutions, you can use the general solution y=C_1y_1+...+C_ny_n and eliminate the constants by successive differentiations. The degree of a differential equation does not necessarily determine its complexity, as there can be multiple ways to form a DE with the same solutions. However, it is important to consider additional solutions that may be introduced when forming a DE.

- #1

epkid08

- 264

- 1

Edit: It has to be homogeneous.

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- #2

kosovtsov

- 70

- 0

(-x+x^2\ln(x))y''= (1+x^2\ln(x))y'-(x+1)y

Formal way to find LODE is as follow.

You can form the "general" solution as

y=C_1y_1+...+C_ny_n

and then eliminate C_1,...C_n by successive differentiations.

- #3

matematikawan

- 338

- 0

ln(x) is a solution for y'=x

Then [tex]y_p=\{\ln(x), e^x\}[/tex]

satisfy the DE (y'-y)(y'-x

- #4

elibj123

- 240

- 2

- #5

matematikawan

- 338

- 0

Is my equation (y'-y)(y'-x-1)= 0 equivalent to (-x+x^2\ln(x))y''= (1+x^2\ln(x))y'-(x+1)y ?

I know a first order nonlinear Riccati DE can be reduced to a linear 2nd order DE. But not particularly sure of this one. Is the degree of a DE has any role to play?

Definition: The degree of a DE is the degree of the highest ordered derivative.

- #6

kosovtsov

- 70

- 0

first order ODE

x*(exp(x)-ln(x))*diff(y(x),x)+(-exp(x)*x+1)*y(x)+(-1+x*ln(x))*exp(x)=0 ,

second order ODE

-ln(x)*x^2*(ln(ln(x))*ln(x)-1)*diff(y(x),x)^2-y(x)*x*(ln(x)+1)*diff(y(x),x)+y(x)*ln(x)*x^2*(ln(ln(x))*ln(x)-1)*diff(y(x),`$`(x,2))+factor(y(x)^2*ln(y(x))*ln(x)+y(x)^2*ln(y(x)))=0 .

- #7

pbandjay

- 118

- 0

Even just looking at the form (y' - y)(y' - x^{-1}) = 0, there are any number of ways to form a differential equation with the given solutions, since D^{n}(e^{x}) = e^{x}, one can use (y^{(n)} - y)(y' - x^{-1}) = 0, for any positive integer n.

Last edited:

- #8

Mute

Homework Helper

- 1,388

- 12

pbandjay said:^{-1}) = 0, there are any number of ways to form a differential equation with the given solutions, since D^{n}(e^{x}) = e^{x}, one can use (y^{(n)}- y)(y' - x^{-1}) = 0, for any positive integer n.

You have to be careful, though. You proposed DE has introduced new solutions: [itex]\exp[\omega_n x][/itex], where [itex]\omega_n[/itex] is an n-th root of 1. So while such a DE does have the desired solutions, it now has many more, which may not be desired.

- #9

matematikawan

- 338

- 0

OK I will take note that 'simplest DE' means linear DE not the lowest order DE.

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model a wide range of phenomena in science and engineering, including motion, growth, and change over time.

Finding the simplest form of a differential equation can make it easier to understand and solve. It also allows for more efficient and accurate calculations, which is crucial in many scientific applications.

The simplicity of a differential equation is determined by the number of terms and operations present in the equation. The fewer terms and operations, the simpler the equation is considered to be.

Using a simple differential equation can help scientists make predictions and better understand the behavior of a system. It also allows for easier integration with other mathematical models and theories.

No, not all differential equations can be simplified into a simpler form. Some equations are inherently complex and cannot be reduced. However, scientists often aim to simplify equations as much as possible to make them more manageable and easier to work with.

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