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consider the following DE

$f''(x) = (1+f'(x)^2)^{\frac{3}{2}}$

given the initial condition $f(0) = f'(0) = 0$ solve for a sol

any hint on how to get started on this?

let $g = f'$

then $\int \dfrac{g'}{(\sqrt{1+g^2})^{\frac{3}{2}}}dt=\int dt$

so I guess it looks something like $x = \sqrt{1+g^2}$, but then it doesn't work out

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    Integration!... – 2017-01-18
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    I have tried but it does not seem to get anywhere – 2017-01-18
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    $f'=u$ so $\int \dfrac{du}{(\sqrt{1+u^2})^3}=\int dx$ – 2017-01-18
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    ok so I have tried it so far, but I think what I have trouble with is the fact that $du$ is the only term in the denominator, so I'm not sure where to go from there – 2017-01-18
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    Where did $dt$ come from in your question? – 2017-01-18
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    Note that the equation is that of a curve with constant curvature, hence a circle of radius $1$. – 2017-01-18

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$y=f(x)$ set $v$ equal to $y'=\frac{dy}{dx}$ and then rewrite the equation as $v'=(1+v^2)^{3/2}$ so now by separation we have $\frac{1}{\left (1+v^2\right )^{3/2}}\text{d}v=\text{d}x$ The right side of this is trivial, as for the left side making the substitution $v=\tan u\to \text{d}v=\sec^2 u\text{d}u\implies \int\frac{\sec^2 u}{\sec^3 u}\text{d}u=\int\cos u\text{d}u=\sin u=\frac{v}{\sqrt{1+v^2}}+C$ So in total we have $\frac{v}{\sqrt{1+v^2}}=x+C$ and $v=y'$ Edit: Forgot about the initial conditions, because $v(0)=f'(0)=0$ then the constant is just 0. Now we can solve for $v$ by squaring both sides and doing some algebra to get that $v^2=\frac{x^2}{1-x^2}\implies\frac{dy}{dx}=\frac{x}{\sqrt{1-x^2}}$ by a u-sub the right integral becomes obvious and we get that $y=-\sqrt{1-x^2}+C$ and from the initial conditions we get that this constant is equal to 1 so... $\boxed{f(x)=1-\sqrt{1-x^2}}$

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    $$\int\dfrac{x}{\sqrt{1-x^2}}dx=-\sqrt{1-x^2}$$ – 2017-01-18
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    @MyGlasses my bad, I will fix that, thank you – 2017-01-18
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It is advantageous to recognize slope $\tan\phi$ as parameter. $$ f^{\prime}== \tan \phi \tag1$$ $$ f^{\prime \prime}== \sec^2 \phi \frac{d\phi}{dx} \tag2 $$ $$ \sec^3 {\phi} = \sec^2 \phi \frac{d\phi}{dx} \tag3 $$ $$ \cos \phi\, d\phi = dx \tag4 $$ $$\sin\phi = x+c = x, \tag5 $$ with given BC. Now letting $y=f(x),$

$$ \frac{dy}{d\phi}= \frac{dy/dx}{d\phi/dx}= \frac{\tan\phi}{1/\cos \phi} = \sin \phi \tag6 $$

Integrating,

$$y = -\cos \phi +c2 = 1-\cos \phi, \tag7 $$

with given BC.

Parametric equations are:

$$ (x,y)= (\sin\phi,1-\cos\phi) \tag8 $$

A circle of unit radius tangential at origin can be readily recognized.

$$ x^2 + (y-1)^2 = 1 \tag9$$ The given ODE is formula of unit curvature.