# Second degree differential equation

• Oct 20th 2011, 05:04 PM
Darkprince
Second degree differential equation
Hello again. How do I solve the diff. equation (dy/dx)^2 + x* (dy/dx) - y = 0?

It's the first time that I face a second degree diff. equation and don't know how to handle it. Any help would be appreciated! Thank you very much!
• Oct 20th 2011, 05:16 PM
chisigma
Re: Second degree differential equation
Quote:

Originally Posted by Darkprince
Hello again. How do I solve the diff. equation (dy/dx)^2 + x* (dy/dx) - y = 0?

It's the first time that I face a second degree diff. equation and don't know how to handle it. Any help would be appreciated! Thank you very much!

Usually as second degree differential equation one means a differential equation in which the second derivative of the unknown function appears... in the differential equation You wrote it does appear the square on the first derivative of the unknown function...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 20th 2011, 05:30 PM
Darkprince
Re: Second degree differential equation
But degree is the order of the highest derivative! When the second derivative appears it is a second order diff. equation not a second degree.
• Oct 20th 2011, 06:54 PM
chisigma
Re: Second degree differential equation
Quote:

Originally Posted by Darkprince
But degree is the order of the highest derivative! When the second derivative appears it is a second order diff. equation not a second degree.

All right!... in that case we write the ODE in the more 'conventional' form...

$\displaystyle y= x\ y^{'} + y^{'\ 2}$ (1)

The approach of (1) is a little 'unusual' ... if we derive both terms respect to x we have...

$\displaystyle y^{'}= y^{'} + (x + 2\ y^{'})\ \frac{d y^{'}}{dx} \implies (x + 2\ y^{'})\ \frac{d y^{'}}{dx} =0$ (2)

... so that we have two possible solutions of (1)...

a) if...

$\displaystyle y^{''}=0$ (3)

... the solution is ...

$\displaystyle y= c\ x + c^{2}$ (4)

b) if...

$\displaystyle y^{'}= - \frac{x}{2}$ (5)

... then setting $\displaystyle y^{'}= t$ and using (1) and the result found in a) we have...

$\displaystyle x=-2\ t$

$\displaystyle y= -t^{2}$ (6)

... which is equivalent to...

$\displaystyle y=-\frac{x^{2}}{4}$ (7)

The solution found in b) is usually called 'singular solution'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 21st 2011, 12:29 AM
Darkprince
Re: Second degree differential equation
Thank you very much! I also have in addition to determine the initial values y(x0) = y0 for which there are no solutions, a unique solution, two or more solutions. How do I do that? Thanks again!
• Oct 21st 2011, 02:58 AM
Ackbeet
Re: Second degree differential equation
Quote:

Originally Posted by Darkprince
Hello again. How do I solve the diff. equation (dy/dx)^2 + x* (dy/dx) - y = 0?

It's the first time that I face a second degree diff. equation and don't know how to handle it. Any help would be appreciated! Thank you very much!

I'm not sure there is clarity here, yet. Note that

$\displaystyle \left(\frac{dy}{dx}\right)^{2}\not=\frac{d^{2}y}{d x^{2}}.$

The first is the square of the first derivative (which is technically what you wrote), and the second is the second derivative. Which is it in your OP?
• Oct 21st 2011, 03:25 AM
Darkprince
Re: Second degree differential equation
Chi Sigma gave me the correct solution, my equation was the first derivative to the square. But let the general solution be y=c*x + c^2 where c is an arbitrary constant, how do we find initial values y(x0) = yo for which there are no solutions , a unique solution or two (or more) solutions? How can we determine for a differential equation how many solutions does it have and in my case for the above diff equation? Thanks again very much!
• Oct 22nd 2011, 12:08 PM
Darkprince
Re: Second degree differential equation
Still cant find initial solutions for the diff equation to have a unique, none, or two (or more) solutions. our general solution is y=c*x+c^2 and any c would give a unique solution. at the same time our singular solution had a factor of x^2 so there is no initial condition for the general solution to equal the singular solution! any help will be greatly appreciated!
• Oct 23rd 2011, 01:14 AM
chisigma
Re: Second degree differential equation
Quote:

Originally Posted by Darkprince
Still cant find initial solutions for the diff equation to have a unique, none, or two (or more) solutions. our general solution is y=c*x+c^2 and any c would give a unique solution. at the same time our singular solution had a factor of x^2 so there is no initial condition for the general solution to equal the singular solution! any help will be greatly appreciated!

I have been away for two days and apologize for that!... If You have an 'initial condition' $\displaystyle y_{0}=y(x_{0})$ , then first You verify that it is compatible with the 'singular solution' $\displaystyle y(x)=-\frac{x^{2}}{4}$, i.e. if $\displaystyle y_{0}= -\frac{x_{0}^{2}}{4}$. If that is true, the the 'singular solution' is one of the solution of the DE. After that You find the values of the constant c that satisfy the second degree equation $\displaystyle c^{2} + c\ x_{0} -y_{0}=0$ , that can have two, one o no real solutions...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 23rd 2011, 05:17 AM
Darkprince
Re: Second degree differential equation
Solving the equation c^2 + c*x0 - y0 yields that the equation has zero solution when (x0)^2+4*y0<0, one solution if (x0)^2+4*y0=0 and two solutions if (x0)^2+4*y0>0.
Can I say anything else about that? Thanks again for your time!

Basically if discriminant is zero we get the real solution, else for <0 we get y0<-x0^2/4 and for >0 we get y0>-x0^2/4.
• Oct 23rd 2011, 12:19 PM
chisigma
Re: Second degree differential equation
Quote:

Originally Posted by Darkprince
Solving the equation c^2 + c*x0 - y0 yields that the equation has zero solution when (x0)^2+4*y0<0, one solution if (x0)^2+4*y0=0 and two solutions if (x0)^2+4*y0>0.
Can I say anything else about that? Thanks again for your time!

Basically if discriminant is zero we get the real solution, else for <0 we get y0<-x0^2/4 and for >0 we get y0>-x0^2/4.

If $\displaystyle x_{0}^{2}+4\ y_{0}=0$ then You have two solution, one of which is the 'singular solution' $\displaystyle y=-\frac{x^{2}}{4}$, so that is...

$\displaystyle x_{0}^{2}+4\ y_{0}<0 \implies \text {no solutions}$

$\displaystyle x_{0}^{2}+4\ y_{0} \ge 0 \implies \text {two solutions}$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Oct 23rd 2011, 04:29 PM
Darkprince
Re: Second degree differential equation
Thanks again for your time!