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How would you go and solve:

$$[f'(x)]^2=a^2\frac{T_2-T_1}{l}f(x)+a^2T_1$$

Where $a$, $l$, $T_1$, $T_2$ are constants and $f(0)=0$?

Here is my try. From the equation we can infer that if $f(x)$ is a polynomial, it must be a second degree polynomial, since its derivative (1st degree) squared is equal to the polynomial itself plus some constant. Since $f(0)=0$, we can write the polynomial as:

$$f(x)=cx^2+dx$$

Substituting this into the first equation whe get that:

$$c=a^2\frac{T_2-T_1}{4l}$$ $$d=a\sqrt{T_1}$$

And thus we found the polynomial. Is this solution acceptable?

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This is not the complete set of solutions.$$\frac{dy}{dx}=\pm\sqrt{a^2\frac{T_2-T_1}ly+a^2T_1}$$is a variable separable ODE assuming $l$ is a constant as well.$$\int\frac{dy}{\sqrt{a^2\frac{T_2-T_1}ly+a^2T_1}}=\pm\int dx$$ The general solution is$$\frac{2l}{a^2(T_2-T_1)}\sqrt{a^2\frac{T_2-T_1}ly+a^2T_1}=\pm x+C$$

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  • $\begingroup$ Thanks. This also means that there are no other solutions, right? $\endgroup$ Jan 23, 2019 at 8:26
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    $\begingroup$ @marcozz Well, the problem is that you can also switch from the "$+$"-solution to the "$-$"-solution at any point you like if you manage to glue them together with suitable constants $C$. This constant $C$ can be different for any of the pieces of your then piecewise defined function. $\endgroup$
    – maxmilgram
    Jan 23, 2019 at 8:50

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