My Math Forum Reduction of second order ode to first order

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 March 30th, 2012, 01:49 AM #1 Member   Joined: Nov 2011 Posts: 40 Thanks: 0 Reduction of second order ode to first order Hey, I've been working on this problem, I use the hint and end up with a seperable equation then get \begin{align} & p=\pm \frac{1}{\sqrt{{{y}^{2}}-2c}} \\ & \text{Then}\,\,\,\text{integrating}\,\,\,\text{aga in (using}\,\,wolframalpha) \\ & y=\pm \log \left( \sqrt{{{y}^{2}}-2c}+y \right)+c \\ & A{{e}^{\pm y}}=\sqrt{{{y}^{2}}-2c}+y \\ \end{align} after getting p is it correct to integrate p to get y? It doesn't look correct in my working but I can't find another way to do it. I thought about trying to solve the equation as y'= +/- ... instead as a first order ode but it seems too complicated, Is there something simple im not seeing? Thanks in advance
 March 30th, 2012, 04:50 AM #2 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: Reduction of second order ode to first order Another way : y' y = -y''/y'² y²/2= (1/y' ) +c Y(0)=1 ; y'(0) = -1 --> 1/2=-1+c ; c=3/2 y²= ('2/y' ) +3 y' = 2/(y²-3) dy/(y²-3) = 2 dx ln[(y-V3)/(y+V3)]/(2V3) = 2x +C V3 = sqrt(3) ln[(1-V3)/(1+V3)] /(2V3)= C ln[(y-V3)/(y+V3)] = 4 V3 x +ln[(1-V3)/(1+V3)] (y-V3)/(y+V3) = [(1-V3)/(1+V3)]*exp((4 V2) x ) y = ...
 March 30th, 2012, 06:24 AM #3 Member   Joined: Nov 2011 Posts: 40 Thanks: 0 Re: Reduction of second order ode to first order I just worked on it a little more and got $\pm dy \sqrt{y^2- 2c}= dx$ Just the p formula I wrote in the first post rearranged \begin{align} & y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1 \\ & \pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides \\ & 1=1-2c \\ & c=0 \\ & So\,y'(x)=\pm \frac{1}{y} \\ & \pm ydy=dx \\ & \pm \frac{1}{2}{{y}^{2}}=x+c \\ & \pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c \\ & \therefore y(x)=\sqrt{2x\pm \frac{1}{2}} \\ \end{align}

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