- #1
amcavoy
- 665
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I cannot get the correct answer to this for some reason:
[tex]t^2y'+2ty-y^3=0[/tex]
I use the substitution [itex]v=y^{1-n}=y^{-2}\implies y=v^{-\frac{1}{2}}[/itex] and come up with [itex]y'=-\frac{1}{2}v^{-\frac{3}{2}}[/itex] and [itex]y^3=v^{-\frac{3}{2}}[/itex].
[tex]-\frac{1}{2}t^2v^{-\frac{3}{2}}+2tv^{-\frac{1}{2}}-v^{-\frac{3}{2}}=0[/tex]
Then multiplying through by [itex]v^{-2}[/itex] gives:
[tex]-\frac{1}{2}t^2v^{3}+2t-v^{3}=0=2t-\left(\frac{1}{2}t^2+1\right)v^3[/tex]
For which I would say:
[tex]v=\left(\frac{2t}{\frac{1}{2}t^2+1}\right)^{\frac{1}{3}}[/tex]
But according to the back of my book, that answer is incorrect. What did I do wrong?
Thanks for your help.
[tex]t^2y'+2ty-y^3=0[/tex]
I use the substitution [itex]v=y^{1-n}=y^{-2}\implies y=v^{-\frac{1}{2}}[/itex] and come up with [itex]y'=-\frac{1}{2}v^{-\frac{3}{2}}[/itex] and [itex]y^3=v^{-\frac{3}{2}}[/itex].
[tex]-\frac{1}{2}t^2v^{-\frac{3}{2}}+2tv^{-\frac{1}{2}}-v^{-\frac{3}{2}}=0[/tex]
Then multiplying through by [itex]v^{-2}[/itex] gives:
[tex]-\frac{1}{2}t^2v^{3}+2t-v^{3}=0=2t-\left(\frac{1}{2}t^2+1\right)v^3[/tex]
For which I would say:
[tex]v=\left(\frac{2t}{\frac{1}{2}t^2+1}\right)^{\frac{1}{3}}[/tex]
But according to the back of my book, that answer is incorrect. What did I do wrong?
Thanks for your help.