Solving Riccatti Equation: 2nd Order Diff. Eq.

[yoyo]

Consider the following Riccatti equation:

dy/dx= -y^2+ a(x)y + b(x) (Eq. 2)

Here a(x) and b(x) are arbitrary functions.

1. Set y(x)= u'(x)/ u(x) where u(x) is a function to be determined. Use (Eq. 2) to show that u(x) satisfies a linear differential equation of second order.

2. Set a=0 and b=1. Solve (Eq. 2) by separating variables using y(0)=0 as the initial condition.

3. Find the function u(x) corresponding to the case a=0, b=1 and y(0)=0. When solving the second order equation for u(x) assume that u'(0)= y(0) and u(0)= 1. Compare the solution of (Eq. 2) from Part 2 with u'(x)/u(x).

4. Eplain why it is nice to replace a non-linear first order differential equation with a linear second order one.
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for part 1 here is what i have substituting y=u'/u

(dy/dx)= -(u'/u)^2+a(x)(u'/u)+b(x)

multiply everything by u to get:

u(dy/dx)=-u(u')^2+a(x)(u')+b(x)u

So am i done here or is there more? (I have a really lousy professor who
just barely got his phD and can't really teach. therefore i am having a lot of diffuculty understanding the matrial. i feel like i am teaching myself ODE...so please help)

 

[dextercioby]

https://www.physicsforums.com/showthread.php?t=65555

Please DO NOT DOUBLE POST...!Especially once you've gotten help in that thread...

Daniel.

 

[saltydog]

multiply everything by u to get:

u(dy/dx)=-u(u')^2+a(x)(u')+b(x)u

So am i done here or is there more? (I have a really lousy professor who
just barely got his phD and can't really teach. therefore i am having a lot of diffuculty understanding the matrial. i feel like i am teaching myself ODE...so please help)

No, wish to rid the equation of y and just have u's. First, just treat a(x) and b(x) as a and b with the understanding that they are functions of x ok. Then we have:

y^{'}=-y^2+ay+b

Letting y=\frac{u^{'}}{u} and substituting this into the ODE, we get:

\frac{uu^{''}-(u^{'})^2}{u^2}=-\frac{(u^{'})^2}{u^2}+\frac{au^'}{u}+b

You understand this part right?

Now, multiplying throghout by u^2 and noting that there is a -(u^{'})^2 on both sides which cancel, we have:

u(u{''})=au(u{'})+bu^2

Divide out the u and place in standard form:

u^{''}-au^{'}-bu=0

That's a good start and it's important to understand this before going further. I assume you'll look at mine, then attempt to go through the steps on your own on paper. Next, do a similar substitution with the following more general form of the Riccati equation:

y^{'}+Qy+Ry^2=P

Using the substitution:

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

With Q,R, and P functions of X so when you're differentiating y remember to differentiate R as well.

 

[saltydog]

Oh and we're not done here: either you or me should/could write a final report on this with at least one plot. Would be better for you if you're taking this in school but if I don't see any follow-up postings by Sun. night, I'll do so (I'm patient).

 

[yoyo]

thanks for the help salty, it really pointed me in the right direction. After some calculus I was able to get u''-au'-bu=0 . however, I still can't figure out number 3, which I've written again below:

Find the function u(x) corresponding to the case a=0, b=1 and y(0)=0. When solving the second order equation for u(x) assume that u'(0)= y(0) and u(0)= 1. Compare the solution of (Eq. 2) from Part 2 with u'(x)/u(x).

I got u''-1=0, and then rewrote it as d^y/dx^2 =1 . From here though I'm stuck.

Also, what is your opinion on number 4? is it because of the properties of a linear equation?

Thanks

 

[saltydog]

Find the function u(x) corresponding to the case a=0, b=1 and y(0)=0. When solving the second order equation for u(x) assume that u'(0)= y(0) and u(0)= 1. Compare the solution of (Eq. 2) from Part 2 with u'(x)/u(x).

I got u''-1=0, and then rewrote it as d^y/dx^2 =1 . From here though I'm stuck.

Also, what is your opinion on number 4? is it because of the properties of a linear equation?

Thanks

Well, if b=1 then the second order eq. in u(x) becomes:

u^{''}-u=0

Solving this the usual way, you know how to right, gets:
u(x)=c_1e^x+c_2e^{-x}

Now, you can plug in u'(0)=u(0)=1 into this and its derivative, get two equations, two unknowns, find c_1 and c_2.

Also, #4: non-linear equations are much harder to solve than linear equations.