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I've been stuck on trying to solve $\dfrac{d^2(y)}{dx^2} = 2y$ for a while now. I tried doing a change of variables (multiplyinh the whole equation by $dx^2$) but then I realized you can't really integrate square variables. So I thought what if I made the substitution $x^2 = z$ and solve $\frac{d^2(y)}{y} = 2dz$?

The question is essentially: Am I allowed to do that? Does this make sense mathematically? If so that would be great as I'd have my answer.

Any help would be much appreciated.

user12345
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user45815
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    Nope, none of that made any sense. This is a second order linear ODE with constant coefficients. There is a standard theory for solving these using the characteristic polynomial (which amounts to a clever "guess-and-check" technique that always works). – Ian Apr 25 '17 at 00:05
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    Have a look at this post: https://math.stackexchange.com/questions/21199/is-frac-textrmdy-textrmdx-not-a-ratio. It provides some useful insights as to what manipulations you can do. – mrnovice Apr 25 '17 at 00:14

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You cannot do this because $$\dfrac{d^2(y)}{dx^2}\neq\dfrac{d^2(y)}{d(x^2)}$$ along with other issues. You CAN rewrite this as $$\dfrac{d}{dx}\dfrac{d}{dx}(y)$$ It's just a notation to write $dx^2$ when what is really meant is more like $(dx)^2$. This differential equation is relatively simple enough to just try some guesses. Try some cosine or sine or exponential functions, etc. In general, you can try $$y=C_1\cos(x)+C_2\sin(x)+C_2e^{C_3x}$$ Or, as the comment said, this is a well-known type of Differential equation

user12345
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  • Thanks I'll try what you suggested. But what if I put (dx)^2 on the RHS and write double integrals on both sides? Doesn't that make sense? – user45815 Apr 25 '17 at 00:19
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    I see what you mean. So, something like $$\int\int\dfrac{1}{2y}dydy=\int\int dxdx$$? I'll let you try that and see if it works (to see if it does, just plug your "solution" for $y$ back into your differential equation and make sure the equation is satisfied). (it shouldn't work). The usual way to do this is to know that the characteristic equation is $r^2-2=0\implies r=\pm\sqrt{2}$ and those $r$ are the exponents to the exponential (i.e. $$y=c_1e^{\sqrt{2}x}+c_2e^{-\sqrt{2}x}$$) – user12345 Apr 25 '17 at 00:23
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    @user45815 The way I got that characteristic can be explained here: http://tutorial.math.lamar.edu/Classes/DE/HOHomogeneousDE.aspx – user12345 Apr 25 '17 at 00:24
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Your ODE can be rewritten as $y''-2y=0$. So you can assoiciate it to the polynomial $p(x)=x^2-2$, which has the roots $\pm \sqrt{2}$. Therefore the general sulution is given by $y(x)=ae^{-\sqrt{2}}+be^{\sqrt{2}}$.

Let me explain why this i true (the following can be done rigorously, but I'll show you the idea behind it, because I think it's easier to remember ideas than general techniques).

Let's have a look at a more general situation. Consider $$y^{(n)}+a_{n-1}y^{(n-1)}+...+a_1y'+a_0y=0.$$ We now want to guess solutions. For this we assume $y$ can be written as $y(x)=e^{\lambda x}$. Observe that in this form $y^{(k)}(x)=\lambda^k e^{\lambda x}$. Plugging this into our equation and dividing by $e^{\lambda x}$ yields $$p(\lambda):= \lambda^{n}+a_{n-1}\lambda^{n-1}+...+a_1\lambda+a_0=0.$$ So you can associate your ODE to a polynomial. If $\lambda_1,...,\lambda_n$ are the (complex) roots of this polynomial, then $y_k(x)=e^{\lambda_k x}$ is a solution for your ODE. Because this type of ODE is linear, the general solution is given by $$y(x)=b_1y_1(x)+...+b_ny_n(x).$$ If one of the rootes is complex (and therefore another one), just take the real and imaginary part of the corresponding $y_k$.

Frieder Jäckel
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