How do you solve this nonlinear first order DE

[kingkongsr.]

I am having trouble solving the following nonlinear first order differential equation:

dy/dx = mx + b - k*y^2

The variables m, b and k are constants.

I have had DE in school, but it was mostly linear first order, so I am not sure how to solve this one. Someone has recommended Riccati but I do not know how to go about solving it.

Please help.

 

[John Creighto]

Aside from describing functions I haven’t studied nonlinear differential equations. It would seem to me though if you differentiate both sides of the equation twice you would get an equation that only depends on y. Perhaps this could be thought as somewhat analogous to the homogenous equation.

 

[kingkongsr.]

How would you go about solving the equation? I need a step by step.

 

[ice109]

$$\left\{\left\{y(x)\to \frac{\frac{k m \text{Bi}'\left(\frac{b k+m x k}{(k m)^{2/3}}\right)}{(k m)^{2/3}}+\frac{k m \text{Ai}'\left(\frac{b k+m x k}{(k m)^{2/3}}\right) c_1}{(k m)^{2/3}}}{k \left(\text{Bi}\left(\frac{b k+m x k}{(k m)^{2/3}}\right)+\text{Ai}\left(\frac{b k+m x k}{(k m)^{2/3}}\right) c_1\right)}\right\}\right\}$$

where Ai is the airy function Ai[x], Bi is again some variant of the airy function Bi[x], and the primes are their respective derivatives

 

[kingkongsr.]

Well that's beyond my understanding.

I was hoping for something easy.

 

[ice109]

what do you mean beyond your understand? it's just a function, put a number in get a number out

 

[kingkongsr.]

What's an airy function?

 

[kingkongsr.]

Show me the process by which you solved it, starting with the original equation. Then explain to me what the airy function is.

 

[ice109]

the process by which i solved it is simple: i plugged it into mathematica. and i have no idea what the airy function is but I'm sure wikipedia does.

 

[kingkongsr.]

I just looked at the equation again, it is suppose to be: dy/dx = m*x + b - k*y^2 where m is multiplied by x. However, can both sides be multiplied by dx such that dy = (m*x + b - k*y^2) dx where the right hand side can be integrated with respect to x. Then both sides are divided by the integral of the right side such that dy/[IntegralWithRespectx(m*x + b - k*y^2)] = 1. Then the left hand side can be integrated with respect to y such that, IntegralWithRespecty[1/[IntegralWithRespectx(m*x + b - k*y^2)]] =1. I have never heard of that, but can that be done?

 

[Marco_84]

I just looked at the equation again, it is suppose to be: dy/dx = m*x + b - k*y^2 where m is multiplied by x. However, can both sides be multiplied by dx such that dy = (m*x + b - k*y^2) dx where the right hand side can be integrated with respect to x. Then both sides are divided by the integral of the right side such that dy/[IntegralWithRespectx(m*x + b - k*y^2)] = 1. Then the left hand side can be integrated with respect to y such that, IntegralWithRespecty[1/[IntegralWithRespectx(m*x + b - k*y^2)]] =1. I have never heard of that, but can that be done?

u don't know y, how can u perform that int?

matco

 

[kingkongsr.]

I just need someone to explain to me what the answer to the equation is. Can someone do that without plugging it into a computer program.

 

[Mute]

I just looked at the equation again, it is suppose to be: dy/dx = m*x + b - k*y^2 where m is multiplied by x. However, can both sides be multiplied by dx such that dy = (m*x + b - k*y^2) dx where the right hand side can be integrated with respect to x. Then both sides are divided by the integral of the right side such that dy/[IntegralWithRespectx(m*x + b - k*y^2)] = 1. Then the left hand side can be integrated with respect to y such that, IntegralWithRespecty[1/[IntegralWithRespectx(m*x + b - k*y^2)]] =1. I have never heard of that, but can that be done?

No, what you're trying to do there is a separation of variables, but you can't separate the y^2 term on the right hand side from the x terms, so you still have to integrate over y^2 with respect to x - effectively what you've done is converted a differential equation into an "integral equation".

It's weird that taking another derivative, as suggested above, doesn't seem to be be useful. Taking the derivative, you get:

$$\frac{d^2y}{dx^2} + 2ky = m$$

This is a linear second order homogeneous equation which can be solved in closed form, but the first derivative of the solution that equation can't be matched to $mx + b$. Huh... I haven't seen that before...

If you do want to try and solve it via the suggested "Riccati Equation", of which this is apparently a form of, with $q_1(x) = 0$, try looking at the wikipedia article: http://en.wikipedia.org/wiki/Riccati_equation. Of course, if computers are giving Airy functions as solutions, then you're probably going to have to resort to a power series solution once you boil that down to the linear 2nd order equation as described in the article (or perhaps you can identify the resulting second order ODE with Airy's equation: $u'' + xu = 0$ - but probably not. I don't know; that computer solution is ugly.)

 

[coomast]

You can rewrite this equation into:

$$\frac{dy}{dx}+ky^2=mx+b$$

Which is a Riccati equation, using the following transformation:

$$y(x)=\frac{1}{ku}\frac{du}{dx}$$

One obtains:

$$\frac{d^2u}{dx^2}-k(mx+b)u=0$$

Substituting now:

$$k(mx+b)=(km)^{2/3}t$$

The equation becomes:

$$\frac{d^2u}{dt^2}-tu=0$$

Which is the differential equation of Airy, with solution:

$$u(t)=c_1 A_i(t)+c_2 B_i(t)$$

In which now Ai and Bi the two independent solutions of the equation, called the functions of Airy. Upon using the inverse transformations, one gets:

$$u(x)=c_1 A_i\left[\frac{k(mx+b)}{(km)^{2/3}}\right]+ c_2 B_i\left[\frac{k(mx+b)}{(km)^{2/3}}\right]$$

The derivative is:

$$\frac{du}{dx}=c_1 \frac{km}{(km)^{2/3}} A_i^{'}\left[\frac{k(mx+b)}{(km)^{2/3}}\right]+ c_2 \frac{km}{(km)^{2/3}} B_i^{'}\left[\frac{k(mx+b)}{(km)^{2/3}}\right]$$

Thus the solution to the original equation becomes (after rewriting a bit):

$$y(x)=\frac{(km)^{2/3}}{k} \frac{A_i^{'}\left[\frac{\displaystyle k(mx+b)}{\displaystyle (km)^{2/3}}\right]+ C B_i^{'}\left[\frac{\displaystyle k(mx+b)}{\displaystyle (km)^{2/3}}\right]} {A_i\left[\frac{\displaystyle k(mx+b)}{\displaystyle (km)^{2/3}}\right]+ C B_i\left[\frac{\displaystyle k(mx+b)}{\displaystyle (km)^{2/3}}\right]}$$

Hope there are no typo's in here.