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June 4th, 2014, 05:59 AM  #1 
Newbie Joined: Jun 2014 From: UK Posts: 1 Thanks: 0  Solving ODEs without making an Ansatz
Hi, Last year in my final year of sixth form (High school?), one of the topics we covered was solving linear ODEs with various methods. Now I'm in University we've recovered solving second order or higher with constant coefficients, but the only method we've used to do so is by making an Ansatz. Last year it was proved to us that the solution must be of the form e^kx with use of differential operators, but I can't find this proof anywhere online, and I was wondering if anyone was able to reproduce it. Thanks. 
June 4th, 2014, 08:59 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra 
I can sketch a proof (due to Tom Apostol) for second order equations. Briefly, we can prove that there is a unique nontrivial solution to the differential equation $$y^{\prime\prime} + by = 0 \qquad \text{with initial conditions} \qquad y(c) = d \; y^\prime(c) = e$$ We do this by assuming that $f(x)$ and $g(x)$ are both solutions to the problem, and then creating the Taylor expansion of $h(x) = f(x)  g(x)$. The error term of the expansion can be made arbitrarily small. We can then observe that solutions to $y^{\prime\prime} + by = 0$ are easy to see. If \begin{align*} b &\lt 0 \qquad \text{write $k^2 = b$ and then} & y &= c_1 e^{kx} + c_2 e^{kx} \\ b &= 0 \qquad \text{then} & y &= c_1 + c_2 x \\ b &\gt 0 \qquad \text{write $k^2 = b$ and then} & y &= c_1 \sin{(kx)} + c_2 \cos{(kx)} \\ \end{align*} We then prove that these represent all solutions, by showing that if $f(x)$ is a solution of the initial value problem described above, and there exist constants $c_1$ and $c_2$ such that $y(0) = f(0)$ and $y^\prime(0) = f^\prime(0)$, then by the uniqueness theorem they are the same. And thus we have all solutions. Finally, we can show that the general differential equation $$y^{\prime\prime} + ay^\prime + by = 0$$ reduces to the special form studied above. If we write $y = uv$ we get $$y^{\prime\prime} + ay^\prime + by = (v^{\prime\prime} + av^\prime + bv)u + (2v^\prime + av)u^\prime + vu^{\prime\prime}$$ We can then choose $v$ so that the coefficient of $u^\prime$ is zero which gives $v = e^\frac{ax}{2}$. Then $$v^{\prime\prime} + av^\prime + bv = \frac{4b a^2}{4}v$$ and $$y^{\prime\prime} + ay^\prime + by = \left( u^{\prime\prime} + \frac{4b a^2}{4}u \right)v$$ But $v$ is never zero, so for $y^{\prime\prime} + ay^\prime + by = 0$ we must have $$u^{\prime\prime} + \frac{4b a^2}{4}u = 0$$ which is an equation of the simpler for discussed above. The second order (and higher) equations have a characteristic polynomial equation, the roots of which determine distinct solutions of the differential equation. All solutions of the equation are linear combination of these distinct solutions. Positive real roots $r = k$ correspond to solutions $c_i e^{kx}$. Complex roots $r = p \pm iq$ correspond to solutions $e^{px}\left( c_i \cos {qx} + c_j \sin {qx} \right)$. When roots are repeated, the corresponding solutions are multiplied by successive powers of $x$, $c_i f(x) + c_j x f(x) + \cdots$. This is probably the quickest way of forming a solution if you can factor the characteristic equation into linear and/or quadratic roots. This paragraph is a summary of the Wikipedia page. 

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