# Nonlinear first order

1. Feb 14, 2006

### danmag

Hi,
As far as I know this is a first order, nonlinear diff eqn with both dependant and independant variables...so it is not solvable??

y'+ay^2 = bx

If anybody knows if there is a method to solving it, please let me know.

Thanks,
danmag

2. Feb 14, 2006

### Haelfix

I think this is a Ricatti diff eqn. Its like the Bernouilli equation, but with nonhomogenous parts (its a bit of a pain if I recall right, you need to do some guessing).

But anyway make the substitution v = 1/y and go from there.

3. Feb 15, 2006

### saltydog

As Haelfix mentioned, it's a Riccati equation of the form:

$$y^{'}+Q(x)y+R(x)y^2=P(x)$$

So use the standard approach:

$$y=\frac{u^{'}}{Ru}$$

to convert it to a second-order ODE with variable coefficients. This then can be solved via power series or just bust it up directly with Airy functions (equal . . . nevermind).

4. Feb 19, 2006

### saltydog

Know what, this turned out to be a quagmire. But very interesting nevertheless. For example:

1. What is the relationship between the power-series solution of the reduced Riccati equation:

$$u^{''}-abxu=0$$

and the solution expressed in terms of Airy functions?

(not easy)

2. Using the solutions above, how then does one proceed to solve an IVP for the original Riccati equation:

$$y^{'}+ay^{2}=bx,\quad y(0)=a$$

(just barely)

3. What is the relationship between the Airy function expressed in terms of a hypergeometric series and the integral expression for the Airy function:

$$Ai(x)=\frac{1}{3^{2/3}\Gamma(2/3)} F_1(2/3,z^3/9)-\frac{z}{3^{1/3}\Gamma(1/3)}F_1(4/3,z^3/9)$$

$$Ai(x)=\frac{1}{\pi}\int_0^{\infty}Cos[t^3/3+xt]dt$$

(not even close)

See . . . quagmire.

Last edited: Feb 20, 2006
5. Feb 20, 2006

### saltydog

More complex contours:

So I reduce the equation down to:

$$u^{''}-axy=0$$

Now, consider the Fourier Transform of u(x):

$$g(\xi)=\mathcal{F}\left\{u\right\}= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u(x)e^{-i\xi x}dx$$

So I'll take the Fourier transform of the DE and noting:

$$\mathcal{F}\left\{xu(x)\right\}=i\frac{dg}{d\xi}$$

I obtain:

$$-\xi^2 g-ia\frac{dg}{d\xi}=0$$

That's a piece of cake to solve and yields:

$$g(\xi)=Ce^{i(\xi^3/(3ai)}$$

Now, inverting the transform:

$$u(x)=\frac{C}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{i(\xi^3/(3a)+\xi x)}d\xi$$

That's close. Looks like I can just split up the integral via Euler's formula to get a Cos and iSin. That might be all there is to it but need to check . . .

Last edited: Feb 20, 2006