Solution to 1st order nonlinear differential equation

gulruozkan

Hi,
I'm a PhD student in Operations Management, and I've stumbled across a differential equation while modeling an OM concept. I was wondering if you could help me with this differential equation, or direct me in a way that would help me solve it.

The equation is:
y'[t]+A[t]*(y[t])^2+B[t]*y[t]+G[t]=0.

As you can see, y, A, B, G are all functions of t. Unfortunately, due to nature of the functions A, B and G, I cannot transform the above equation into y'[t]+(y+J[t])^2=0, which would help me use substitution and solve the above equation. If I can get y[t] solution as a function of A[t}, B[t] and G[t], it would really help me with my research.

Thanks a lot for your help in advance.
Best,

Gulru

lzkelley

I'm by no means an expert of DE's, but the format of your equation looks like it might lend itself to a power series substitution method - it would be a certain amount of brute force, but since you're interested in applications it might suffice.

Mute

The nonlinearity in the problem makes a power series solution very difficult, since you have to multiply series together for the y^2 term. It would really only be plausible if you were sure you could neglect terms and retained only a few terms in the series.

Anywho, fortunately you probably don't quite need to resort to that yet, as this equation is of the form of a Riccati equation:

http://en.wikipedia.org/wiki/Riccati_equation

drishtysingh

hello dear.

Ur problem seems to be a non linear homogeneous equation.

Well the 'dsolve' function in matlab usually can solve such equations analytically.

Ive tried to solve an example of your equation and here is the result:

you might, hopefully, solve yours in a similar way.

Hoping that the below will be of use:


dsolve('Df =(cos(t)*(sin(t))^2 + (2*sin(t)*exp(t))+ 2*cos(2*t))','f(0)=0')

ans =

1/4*sin(t)-1/12*sin(3*t)-exp(t)*cos(t)+sin(t)*exp(t)+sin(2*t)+1


Drishtysingh

gulruozkan

Thanks a lot for the replies, got to work around the problem with a less complicated model, but your replies have been very helpful for solving other problems.

cronxeh

It does not get less complicated than this in DE.. you are lucky to even have an analytical solution

brahman

Could you set the problem more completely ? domain, initial condition , ... where are they ?

kosovtsov

The Riccati equation in general form can not be solved. But if you know one particular solution to this ODE, you then can obtain its general solution.

If you consider the functions A,B,G as arbitrary functions, then you can rewrite the ODE in following equivalent form (it's Maple kode)

diff(y(t),t)+A(t)*y(t)^2+B(t)*y(t)-diff(g(t),t)-A(t)*g(t)^2-B(t)*g(t) = 0

that is here G(t)=-diff(g(t),t)-A(t)*g(t)^2-B(t)*g(t)

where now g is a new arbitrary function, then

particular solution is y(t)=g(t) and the general solution (with new arbitrary functions A,B,g) is

y(t) = g(t)+exp(Int(-2*A(t)*g(t)-B(t),t))/(Int(exp(Int(-2*A(t)*g(t)-B(t),t))*A(t),t)+_C1)

Matthew Rodman

I see what you're saying -- and you can also repeat this process and provide a sort of superposition of these solutions, for example:

For any solution, [tex]v_0[/tex], of the Ricatti equation

[tex]v^{\prime } + v^2 + \Psi = 0[/tex]

we can show, through differentiation, that there will always be another
solution of ther form

[tex]
v_1 = v_0 + \frac{e^{-2\int{v_0 dt}}}
{\int{e^{-2\int{v_0 dt}} dt}}
[/tex]

We can now extend this process to an infinite number of solutions, namely,

[tex]v_{j+1} = v_j + \frac{e^{-2\int{v_j dt}}}
{\int{e^{-2\int{v_j dt}} dt}}
[/tex]

Now, it would be helpful to provide a sort of superposition of all these solutions, but
as the Ricatti equation, is non-linear, the principal of superposition does not
directly apply. However, we can easily translate it into a linear equation, namely make the
substitution

[tex]v = \frac{u^{\prime}}{u}
[/tex]

to obtain the second-order (linear) equation,

[tex]u^{\prime \prime} + \Psi u = 0
[/tex]

So, if [tex]u[/tex] may be composed of a superposition of individual solutions, we can then re-translate
this superposition back to the original Ricatti equation. If we have a solution
of the above of

[tex]
u = \sum_j \alpha_j u_j
[/tex]

we can use [tex]v = u^{\prime}/u[/tex] to express it in terms of [tex]v[/tex],

[tex]
u = \sum_j \alpha_j e^{\int{ v_j dt}}
[/tex]

hence we can give a solution of the Ricatti as a superposition of solutions as

[tex]
v = \frac{u^{\prime}}{u} = \frac{d}{dt} \ln{\sum_j \alpha_j e^{\int{ v_j dt}}}
[/tex]

where [tex]v_0[/tex] is the one known solution.

ceramica

hello, all.. i'm very sorry for interupting..

Does anyone can help me to solve this second order non linear ODE:

y'' + (2/x)(y') - (1/2y)(y')(y') = K,

y' = dy/dx

y'' = dy'/dx

y = y(x)

I've already guess y=Ax^2 satisfy this equation, but I want to solve it analitically..

Please help!
Thank before..

gayatriv

Hi

This is Gayatri. I was searching for a method to solve nonlinear differential equations when I came across this thread.

I am stuck with one of the nonlinear differential equations which I have not been able to solve analytically or numerically.It will be of great help if someone can help me in this regard.

The equation which I have is :
y'[x]^2 == (y[x]^2 - c^2) - ((y[x]^2*p^2)*(y[x]^2 - 2*c^2))

Here p and c are constants. The boundary conditions are
y[0]=x0^2
y[L]=c^2

Some one please help me with this solution..

Thanks a lot in advance.

Gayatri

sahararbabi

hello all,
i'm a master student,my major is physics..i have problem in solving a non-linear Riccati differential equation.how can i solve it?
w^2-dw/dr=A(V1-V2)+B+C(V1+V2)
plz help me,
thanks,
sahar

zsviben

Hi,

if the right side of equ is constant,
you can solve it easily
w = k Tan[k (r+c)]

where
k = Sqrt[A(V1-V2)+B+C(V1+V2)]
and
c is the constant of integration what is determined from initial conditions

Bye,

bhatmuzzamil

hi i am phd student in physics,i am finding difficult to solve the first ordre nonlinear differential equation as dy/dx = y^2 - 2x^2 - xc /(c - x)y plz give me some tips in this regard

JJacquelin

HI !

The Riccati ODE can be transformed to a linear ODE, even without knowing a particular solution (see in attachment)

Attached Thumbnails