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From the Question $y''=1+(y')^2$

I used Reduction of Order by taking $y'=z$ to get,

$$z\left(\frac{dz}{dy}\right)=1+z^2$$

Solve the equation by Separation of Variable, to get

$$\frac{1}{2} \ln(1+z^2) = y+c $$

I tried to substitute z=y' back to the equation above and solve the equation by Integration, but that was far very complex, so I got struck here. Please help

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    When $\frac{dy}{dx} = z$, then $\frac{d^2y}{dx^2} = \frac{dz}{dx}$. – 2017-01-12
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    You could still easily go the next step $1+y'^2=1+z^2=C·e^{2y}$, after that this way really gets complicated. – 2017-01-12

2 Answers 2

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We want to solve $$y''=1+(y')^2$$

The usual thing is to replace $y'=v$ and therefore we get

$$v'=1+v^2.$$

This ODE can be solved by the usual methods, just use what you know. For example separation of variables i.e. solve for $v$

$$\int \frac{dv}{1+v^2} = \int dx$$

We get $v(x)=\tan(x+C)$ where $C$ is some constant. Now resubstitute i.e. $$y'=v=\tan(x+C)$$

Just integrate both sides with respect to $x$ i.e.

$$y(x)=\int \tan(x+C) \ dx$$

This can be solved easily, for example use substitution $s=\cos(x+C)$

$$\int \tan(x+C) \ dx=-\int \frac{ds}{s}=-\log(s)=-\log(\cos(x+C))$$

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One thing to recognize is that the derivative of $\tan^{-1}$:

$$ \frac{d}{dx} \tan^{-1}(x) \;\; =\;\; \frac{1}{1+x^2}. $$

We see here that we can rewrite our differential equation as

$$ \frac{dy'}{1 + (y')^2} \;\; =\;\; dx $$

which can be solved as

$$ \tan^{-1}\left (y' \right ) \;\; =\;\; x + c. $$

The tricky part now integrating $y'(x) = \tan(x+c)$. Are you allowed to look this up?

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    I didn't realize the tan^-1 can be used there!! Got it now thanks very much – 2017-01-12