- #1

franklampard8

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$$\frac{\mathrm{d}y(x)/\mathrm{d}x}{\mathrm{d}z(x)/\mathrm{d}x} = a[1-y(x)-z(x)] + b$$

where a, b are constants. Also,

$$y(0) = z(0) = 0$$

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- Thread starter franklampard8
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In summary, the conversation discussed the process of solving a differential equation of the form (dy/dx)/(dz/dx) = a[1-y(x)-z(x)] + b, where a and b are constants. It was mentioned that with only one equation and two unknown functions, there are an infinite number of possible solutions. The differential equation allows for the expression of a relationship between the two functions, y and z. The process of obtaining y = y(z) was also discussed, and it was shown that any derivable function z(x) can be chosen to find the related function y(x). The conversation also touched on the possibility of originally having two equations and using a "trick" to end up with one differential equation

- #1

franklampard8

- 3

- 0

$$\frac{\mathrm{d}y(x)/\mathrm{d}x}{\mathrm{d}z(x)/\mathrm{d}x} = a[1-y(x)-z(x)] + b$$

where a, b are constants. Also,

$$y(0) = z(0) = 0$$

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- #2

JJacquelin

- 801

- 35

franklampard8 said:

$$\frac{\mathrm{d}y(x)/\mathrm{d}x}{\mathrm{d}z(x)/\mathrm{d}x} = a[1-y(x)-z(x)] + b$$

where a, b are constants. Also,

$$y(0) = z(0) = 0$$

Since you have only one equation, but two unknown functions y(x) and z(x) the problem remains undefined. You cannot obtain one function y(x) and one function z(x), but an infinity of couples.

Nevertheless, tne differential equation allows to express a relationship beteween each couple y(x) and z(x). The relationship is on the form y=function of z.

Now, how to obtain y = y(z) ?

dy/dz = a(-y-z)+b

(dy/dz)+a*y = b -a*z

(dY/dz)+a*Y = 0 --> Y = c*exp(-a*z)

c --> f(z)

y = f(z)*exp(-a*z)

dy/dz = (df/dz)exp(-a*z)-a*f*exp(-a*z)

(df/dz)exp(-a*z)-a*f*exp(-a*z) +a*f*exp(-a*z) = b -a*z

(df/dz)exp(-a*z) = b -a*z

df = (b-a*z)exp(a*z)*dz

f = (1/a)*(-a*z+b+1)*exp(a*z)+C

y = (1/a)*(-a*z+b+1) +C*exp(-a*z)

y(0)=z(0)=0 --> 0 = (1/a)*(b+1) +C --> C = -(b+1)/a

y = (1/a)*(-a*z+b+1) -((b+1)/a)*exp(-a*z)

if you want, you can chose any derivable function z(x). Then, the related function y(x) is :

y(x) = (1/a)*(-a*z(x)+b+1) -((b+1)/a)*exp(-a*z(x))

This proves that, due to the incomplete wording, there are an infinity of solutions.

Last edited:

- #3

bigfooted

Gold Member

- 683

- 214

You would then probably get an ode like dy/dz = a(1-y-z)+b which is linear.

The Quotient of First Order Ordinary Derivatives is a mathematical concept that represents the ratio of the derivatives of two functions. It is defined as the derivative of the first function divided by the derivative of the second function.

The Quotient of First Order Ordinary Derivatives is calculated by taking the derivative of the first function and dividing it by the derivative of the second function. This can also be written as (dy/dx)/(du/dx) where y and u are the two functions.

The Quotient of First Order Ordinary Derivatives is important in calculus because it allows us to calculate the rate of change of one function with respect to another. It also helps in solving problems involving optimization and related rates.

Yes, the Quotient of First Order Ordinary Derivatives can be negative. This means that the two functions are changing in opposite directions, resulting in a negative ratio. It is important to consider both the magnitude and sign of the Quotient in interpreting its meaning.

Yes, there are some limitations to using the Quotient of First Order Ordinary Derivatives. It is only applicable when both functions have a defined derivative and when the denominator is not equal to zero. It also does not take into account the higher order derivatives of the functions.

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