# Finding 2nd solution of second order ODE

Function $x_1(t)=e^t$ is a solution of: $$tx''-(2t+1)x'+(t+1)x=0$$ Find second lineary independent solution.

I tried to do it the usual way, substituting $x=e^{\alpha t}$. But that gives me the only one solution that has been given already. I know the second solution will probably be $te^t$, but I dont know how to prove.it

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Your guess $g(t)=t$ does not work, but it's not very far off. I hope my answer below helps you find the solution. – Joonas Ilmavirta Aug 18 '14 at 15:44

Function $x_1(t)=e^t$ is a solution of: $$tx''-(2t+1)x'+(t+1)x=0$$

Let us try the reduction of order approach, which is to try a second solution $x_2(t)=g(t)x_1(t)$. Inserting into the original DE, we have

$$t(g(t)x_1(t))''-(2t+1)(g(t)x_1(t))'+(t+1)(g(t)x_1(t))=0$$ The derivatives are $$(g(t)x_1(t))' = g'(t)x_1(t) + x_1'(t)g(t)$$ $$(g(t)x_1(t))'' = g''(t)x_1(t) + 2g'(t)x_1'(t) + x_1''(t)g(t)$$ Substituting gives $$g(t)\left(tx_1''(t)-(2t+1)x_1'(t)+(t+1)x_1(t)\right) + F(x_1,g,t) =0$$ where $$F(x_1,g,t) = t(g''(t)x_1(t) + 2g'(t)x_1'(t))-(2t+1)g'(t)x_1(t)$$ The term in parentheses two lines above is zero because $x_1$ satisfies the original DE, leaving $$F(x_1,g,t)=0$$ $$t(g''(t)x_1(t) + 2g'(t)x_1'(t))-(2t+1)g'(t)x_1(t) = 0$$ Now since, $x_1=x_1'=e^t$, that term cancels everywhere, leaving $$t(g''(t) + 2g'(t))-(2t+1)g'(t) = 0$$ There are two $2tg'$ terms that also cancel, leaving $$tg''(t)-g'(t) = 0$$ Letting $h=g'$ reduces this to a first order DE: $$th'(t)-h(t) = 0$$ Separation gives $$\frac{1}{h(t)}\frac{dh}{dt} = \frac{1}{t}$$ Now integrating $dt$ $$\int \frac{1}{h} dh = \int \frac{1}{t} dt,$$ $$\log(h) = \log(t)+c,$$ for some constant $c$. Exponentiating both sides gives $$h = Ct,$$ For some other constant $C=e^c$. Since $g'=h$, we integrate $dt$ once more to get $g$ $$g = At^2+B,$$ where $A=\frac{C}{2}$ and $B$ is another constant of integration. Now we put together our final solution $$x_2(t) = (At^2+B)e^t$$ Plugging this into the DE verifies that it is a second solution. You can verify it's linear independence using the Wronkskian determinant: $$\left| \begin{array}{cc} e^t & e^t \left(A t^2+B\right) \\ e^t & e^t \left(A t^2+B\right)+2 At e^t \\ \end{array} \right| = 2 A t e^{2 t},$$ which does not identically vanish on any interval; so the solutions are independent.

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to avoid possible confusions, let me just add that this method is known as "reduction of order". – etothepitimesi Aug 18 '14 at 23:17
Thanks for the clarification. I believe I was taught the incorrect terminology. I have fixed it above. – rajb245 Aug 19 '14 at 18:02

If you have found the solutions $Cf(t)$, a good attempt to find more solutions is to vary the constant by looking at functions of the form $g(t)f(t)$ and solving for $g$. When simplifying the equation for $g$, many terms will disappear since $f$ is a solution.

In your case, try to find a solution of the form $g(t)e^t$ and see what you get for $g$.

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Thank you, right now im trying to solve it this way. – Lugi Aug 18 '14 at 15:46
I got $tf''(t)=f'(t)$. How do I take the antiderivative? – Lugi Aug 18 '14 at 15:56
That's correct. If you define $h=f'$, you get $th'(t)=h(t)$, which should be easier to solve. Then integrate $h$ to get $f$. – Joonas Ilmavirta Aug 18 '14 at 15:58
$f(t)=C_1*\frac{t^2}{2}+C_2$ Is that correct? – Lugi Aug 18 '14 at 16:09
I think this should be the accepted answer, since it came first and the OP was able to solve his problem with its help. Nevertheless, it's just an opinion. – Dmoreno Aug 18 '14 at 16:40

$$tx''-(2t+1)x'+(t+1)x=t(x'-x)'-(t+1)(x'-x)=0$$ $$ty'-(t+1)y=0$$ $$y'-(1+\frac1t)y=0$$ $$e^{\int-1-\frac1tdt}=e^{-t-lnt}=\frac1{te^t}$$ $$\frac{y'}{te^t}-\frac{t+1}{t^2e^t}y=(\frac y{te^t})'=0$$ $$y=ate^t$$ $$x'-x=ate^t$$ $$e^{-t}x'-e^{-t}x=(e^{-t}x)'=at$$ $$\frac x{e^t}=a_2t^2+b$$ $$x=a_2t^2e^t+be^t$$

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