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Consider the differential equation $x''-(2t/1-t^2)x'+(2/1-t^2)x=0$ for $t$ in $(-1,1)$.

Show that $x(t)=t$ is a solution and find a second linearly independent solution $y(t)$.

So I get how to solve the equation and that $t$ is a solution, but how do I find another linearly independant solution?

Any help or suggestions much appreciated. I'm new here so sorry if I have done anything wrong!

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  • $\begingroup$ I think you have a typo there, because $x(t)=t$ is not a solution. $\endgroup$ – gev Mar 20 '13 at 12:28
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    $\begingroup$ @gev, I think it's meant to be read as $$x''-{2t\over1-t^2}x'+{2\over1-t^2}x=0$$, so $x=t$ is a solution. $\endgroup$ – Gerry Myerson Mar 20 '13 at 12:33
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First, your equation I suppose in the shape of: $$x^{''}-{{2t}\over{1−t^2}}x^{'} +{{2}\over{1−t^2}}x=0$$ Your first solution is $x(t) = t$. So $x^{'}(t) = 1$ and $x^{''}(t) = 0$. Substituting into left hand part of the equation we get $0-{{2t}\over{1−t^2}}\cdot{1} +{{2}\over{1−t^2}}t = {{-2t+2t}\over{1-t^2}} = 0$ and this equals to right hand part of the equation, this means also {0}. So solution x(t) = t is one solution of the equation.
The second solution of the equation we assume in the shape of $x(t) = e^{kt}$, where k is value independent on t. So $x^{'}(t) = k{e^{kt}}$ and $x^{''}(t) = k^2{e^{kt}}$. Substituting into the equation we get: $k^2{e^{kt}}-{{2t}\over{1−t^2}}k{e^{kt}} +{{2}\over{1−t^2}}e^{kt}=0$. We need to identify {k} value. Because e^{kt} is not $0$ we can modify the form of the equation to $k^2-{{2t}\over{1−t^2}}k +{{2}\over{1−t^2}}=0$ and we now have for {k} the quadratic equation. Solving this quadratic equation we get $k_{1,2} = {{t\pm\sqrt{3t^2-2}}\over{1-t^2}}$.
So now we have two more terms which satisfy the equation: $$x(t)= {{t\pm\sqrt{3t^2-2}}\over{1-t^2}}$$(one with the $+$ sign in exponent a second with $-$ sign.
Whether this solution meets with everyone again by inserting:
${(t\pm\sqrt{3t^2-2})^2\over{{1-t^2}^2}}-{{2t}\over{1-t^2}}{{t\pm\sqrt{3t^2-2}}\over{1-t^2}}+{{2}\over{1-t^2}} = {{t^2\pm{2t\sqrt{3t^-2}}+3t^2-2-2t^2\mp{2t\sqrt{3t^-2}}+2-2t^2}\over{1-t^2}}={{0}\over{1-t^2}} = 0$. Is also equals to right hand part of equation. So both these solution also complies with the equation.
Including your solution we have three possible $x$: $$x(t) = t$$ $$x(t)= {{t+\sqrt{3t^2-2}}\over{1-t^2}}$$ $$x(t)= {{t-\sqrt{3t^2-2}}\over{1-t^2}}$$

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  • $\begingroup$ amazing thank you SO much :) $\endgroup$ – user67411 Mar 20 '13 at 14:44
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There's a method for doing this, called reduction of order. I have linked to the Wikipedia article, which should get you started.

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Hint: Assume the second solution is $ x_2(t)= u(t)x_1(t) = tu(t)$ and subs back in the ode to find $u(t)$. See here.

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  • $\begingroup$ thats cool i thought this would be the way! Think i can solve the equation now! how to i find another linearly independant solution, is there a method for this? many thanks for your answer! $\endgroup$ – user67411 Mar 20 '13 at 12:41
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    $\begingroup$ @user67411: You are welcome. Note that, your ode is a linear second order ode which, by the theory, should have two linearly independent solutions. $\endgroup$ – Mhenni Benghorbal Mar 20 '13 at 14:26

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