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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 1st, 2017, 05:42 AM #1 Newbie   Joined: Jun 2017 From: South Africa Posts: 1 Thanks: 0 Ode Hi everyone, I have been trying to solve this equation, but to no avail. Can you help me? y''+y'/x+y/x^2=0 Last edited by skipjack; June 1st, 2017 at 07:12 AM.
 June 1st, 2017, 07:11 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,474 Thanks: 2039 Let's assume that x > 0. Substitute x = e^u. The resulting equation is easy to solve.
 June 15th, 2017, 05:18 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Not with those "x"s there, it isn't. The first thing I would do is multiply by $\displaystyle x^2$ so we have $\displaystyle x^2y''+ xy'+ y= 0$, an "Euler type" or "equi-potential" equation. Now, try something of the form $\displaystyle y= x^s$. Then $\displaystyle y'= sx^{s-1}$ and $\displaystyle y''= s(s- 1)x^{s- 1}$ so the equation becomes $\displaystyle s(s- 1)x^s+ sx^s+ x^s= (s(s- 1)+ s+ 1)x^s= 0$. $\displaystyle x^s$ is not always 0 so we must have $\displaystyle s(s- 1)+ s+ 1= s^2+ 1= 0$ which has roots s= i and s= -i. So the general solution to the differential equation is $\displaystyle = Ax^i+ Bx^{-i}$. To get that in terms of real numbers only, write $\displaystyle x^i= e^{\ln(x^i)}= e^{i\ln(x)}$ and use the fact that $\displaystyle e^{ia}= \cos(a)+ i\sin(a)$ to get $\displaystyle x^{i}= \cos(\ln(x))+ i \sin(\ln(x))$. Similarly, $\displaystyle x^{-i}= \cos(-\ln(x))+ i \sin(-\ln(x))= \cos(\ln(x))- \sin(\ln(x))$. We can absorb the "i" into the constants, A and B, and write the general solution as $\displaystyle y= A\cos(\ln(x))+ B \sin(\ln(x))$. (Another way to approach this is to use the fact that the substitution t= ln(x) reduces it to an equivalent equation with constant coefficients. If t = ln(x), then $\displaystyle \frac{dy}{dx}= \frac{dy}{dt}\frac{dt}{dx}= \frac{1}{x}\frac{dy}{dt}$ and $\displaystyle \frac{d^2y}{dx^2}= -\frac{1}{x^2}\frac{dy}{dt}+ \frac{1}{x^2}\frac{dy}{dt}$ so that the differential equation becomes $\displaystyle \frac{d^2y}{dt^2}+ y= 0$.) Last edited by skipjack; June 15th, 2017 at 05:39 AM.
 June 15th, 2017, 05:35 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,474 Thanks: 2039 Why disagree with substituting x = e^u, yet demonstrate how easily the substitution t = ln(x) works?

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