Help! I'm just starting this class and I have no idea what's going on. What I don't understand is, what answer are you supposed to give? My question says "Find the general solution and also the singular solution, if it exists". What the hell does that mean?(adsbygoogle = window.adsbygoogle || []).push({});

Can someone tell me if this is right?

Question: [tex]x(y')^2 - (2x + 3y)y' + 6y = 0[/tex]

so let [tex]p = y'[/tex]

[tex]xp^2 - (2x + 3y)p + 6y = 0[/tex]

[tex]xp^2 - 2xp - 3yp + 6y = 0[/tex]

[tex]-3yp + 6y = -xp^2 + 2xp[/tex]

[tex]y = \frac{(-xp)(p-2)}{(-3)(p-2)}[/tex]

[tex]y = \frac{xp}{3} [/tex]

so then,

[tex]3y' = 3p = xp' + p[/tex]

[tex]2p = xp'[/tex]

[tex]2p = x \frac{dp}{dx}[/tex]

[tex]\frac{dp}{p} = 2\frac{dx}{x}[/tex]

[tex]\ln{P} = 2\ln{x} + \ln{C}[/tex]

[tex]p = cx^2[/tex]

[tex]y' = cx^2[/tex]

[tex]y = cx^3 + b[/tex], where c and b are constants

and that's as far as I got...now what?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Reducing 2nd order to 1st order type question

**Physics Forums | Science Articles, Homework Help, Discussion**