- #1
Shinjo
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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?
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?
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?
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