1. ## Second-Order Nonlinear DE

Can someone please explain to me how to solve this equation:
x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance

2. Originally Posted by AkilMAI
Can someone please explain to me how to solve this equation:
x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
I think you have a typo

$\displaystyle y=ax+b$

$\displaystyle y'=a$

$\displaystyle y''=0$

If you put this into the ODE you get

$\displaystyle x\cdot (0)-(ax+b)\cdot (a)=2 \iff -a^2x-ab=2$

This implies that

$\displaystyle a=0$

but that makes the equation false!

3. Originally Posted by AkilMAI
Can someone please explain to me how to solve this equation:
x*y''-y*y'=2... how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
Considering that is $\displaystyle y(2)=3$ and setting $\displaystyle z=y'$ the DE is reduced to...

$\displaystyle z'= \frac{3}{2}\ z +1$ (1)

... which is a linear first order DE and is solved in standard fashion if You know the 'initial value' of y'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Originally Posted by chisigma
Considering that is $\displaystyle y(2)=3$ and setting $\displaystyle z=y'$ the DE is reduced to...

$\displaystyle z'= \frac{3}{2}\ z +1$ (1)

... which is a linear first order DE and is solved in standard fashion if You know the 'initial value' of y'...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
Please share - how did you arrive at this?

5. Originally Posted by Danny
Please share - how did you arrive at this?
Probably I entered a wrong way...

... my reasoning was that the in the DE...

$\displaystyle x\ y'' - y\ y'=2$ (1)

... with 'initial condition' $\displaystyle y(2)=3$ was possible the 'direct substitution' $\displaystyle x=2, y=3$ so that it becomes...

$\displaystyle 2\ y'' - 3\ y' =2$

... but of course it's a wrong way ...

I had to remember that a second order non linear DE like (1), where all the variables x,y,y',y'' appear, is in general hard to attack...

Sorry for the waste if time I caused...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. Not a waste of time - I just thought you had some cool magic!

7. Originally Posted by AkilMAI
Can someone please explain to me how to solve this equation:
x*y''-y*y'=2...where I mus find the condition that the constants a and b must satisfy for y=ax+b to be a solution.Also how can I find the two solutions that satisfy y=3 when x=2.Thanks in advance
This may help a bit but I can't see how to finish

$\displaystyle xy''-y\cdot y'=2$

Notice that

$\displaystyle \frac{d}{dx}\left( xy'\right)=xy''+y' \iff \frac{d}{dx}\left(xy'-y \right)=xy''$

and

$\displaystyle y\cdot y'=\frac{1}{2}\frac{d}{dx}y^2$

So we can rewrite the ODE as

$\displaystyle \frac{d}{dx}\left( xy'-y-\frac{1}{2}y^2\right)=2$

So we can integrate once to get

$\displaystyle xy'-y-\frac{1}{2}y^2=2x+C$

I can't see any obvious way to solve this.

8. The ODE is now Ricatti. Note also that $\displaystyle y = u-1$ then we have

$\displaystyle 2x u' = u^2 + 4x + 2c-1$

Further if we let

$\displaystyle u = -2x \frac{v'}{v}$then the ODE becomes linear

$\displaystyle 4x^2 v'' + 4xv' + (4x+2c-1)v = 0$!

9. At the OP. Check on the ODE itself. Your question suggests that there is a solution of the form

$\displaystyle y = ax + b$ and as TheEmptySet points out - there's a contradiction.

10. Another possible 'attack' to the DE...

$\displaystyle x\ y'' - y\ y'=2$ (1)

... consists in supposing that y is analytic in x=2 so that is...

$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (2)}{n!}\ (x-2)^{n}= \sum_{n=0}^{\infty} a_{n}\ (x-2)^{n}$ (2)

... and that we know $\displaystyle a_{0}= y(2)$ and $\displaystyle a_{1}= y'(2)$. From (1) we derive first ...

$\displaystyle 2\ y''(2)- y(2)\ y'(2) = 2 \implies y''(2)= 1+\frac{a_{0}\ a_{1}}{2} \implies a_{2}= \frac{1}{2}+ \frac{a_{0}\ a_{1}}{4}$ (2)

... and then...

$\displaystyle 2\ y'''(2) + y''(2) -y(2)\ y''(2) - y'^{2}(2)=0$ (3)

... that permits us to derive $\displaystyle y'''(2)$ and then $\displaystyle a_{3}= \frac{y'''(2)}{6}$... and then all the derivatives of y in x=2 we want...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$