# Just another ordinary differential equation

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?

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$$\frac{2l}{a^2(T_2-T_1)}\sqrt{a^2\frac{T_2-T_1}ly+a^2T_1}=\pm x+C$$
• @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. Jan 23, 2019 at 8:50