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Given this first-order linear equation: $y' ={ {x^2+xy+y^2} \over x^2}$!

I have to show first that it is homogeneous. I divide numerator and denominator of RHS by $x^2$ to get

$dy/dx=1+{y \over x}+({y \over x})^2$

I then do the regular business with v-substitution until I get:

$Integral{1 \over x}dx$ = $Integral(1+v^2)dv$

After Integration and a some simplification I get:

$v^3+3v = 3ln(x)-3c$.

I don't know how to proceed from here.

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I think you mean $y^2$ instead of $\frac{y^2}{x^2}$ (In the title) –  Ishan Banerjee Feb 20 '13 at 5:56
    
@IshanBanerjee thanks, fixed it. –  Blue Pony Inc. Feb 20 '13 at 6:04

1 Answer 1

up vote 1 down vote accepted

Something went wrong after the substitution. I get $$\frac{dv}{1+v^2}=\frac{dx}{x}.$$ On the left we get an arctan.

Remark: In this case, we can get $v$, and therefore $y$, explicitly in terms of $x$. However, in general, when we separate variables to solve the differential equation $\frac{dy}{dx}=f(x)g(y)$, then, even when the integrations are doable, we may not be able to then solve for $y$ explicitly in terms of $x$.

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Thanks, I will double check my substitution. Arctan sounds correct. –  Blue Pony Inc. Feb 20 '13 at 5:58
    
Andre do I substitute y/x with v in the new rewritten equation, or instead substitute y with vx in the original equation on the very first line? I think i misunderstood this whole chapter on homogeneous equations. –  Blue Pony Inc. Feb 20 '13 at 6:00
1  
When you solve, you get something like $\arctan v=\ln(|x|)+C$. It would be nice if there were an initial condition. Anyway, $v=\tan(\ln(|x|)+C)$. So $y=x\tan(\ln(|x|)+C)$. –  André Nicolas Feb 20 '13 at 6:06
    
@Blue Pony You can do both. –  Ishan Banerjee Feb 20 '13 at 6:06
    
Thanks, again, I was able to produce the result with arctan. My substitution process was incorrect. –  Blue Pony Inc. Feb 20 '13 at 6:30

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