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$$\dfrac{d^2y}{dx^2}=\left( \dfrac{dy}{dx} \right)^2 + 1 \quad \text{and} \quad y(0)=\dfrac{dy}{dx}(0)=0 \quad \text{on} \quad \left( \dfrac{\pi}{2},-\dfrac{\pi}{2}\right)$$


Let's define $u=\frac{dy}{dx} \implies \frac{du}{dx}=\frac{d^2y}{dx^2}$. $$\frac{du}{dx} = u^2+1 \implies \int \frac{1}{u^2+1}\;du = \int dx$$

Thus, $$\arctan{u} = x + K_1 \iff u = \tan{(x + K_1)} $$

We integrate again, $$\int \frac{dy}{dx} \; dx = \int \tan{(x + K_1)} \; dx \implies y = \ln{(\cos{(x + K_1)})} + K_2$$

How can I evaluate the particular solution?

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    $\begingroup$ $\tan K_1=0$ and so ... $\endgroup$ – John B Apr 26 '16 at 0:54
  • $\begingroup$ How can you know that $\tan{(K_1)}=0$? $\endgroup$ – hlapointe Apr 26 '16 at 0:58
  • $\begingroup$ compute $y'(x)$ and take $x=0$. $\endgroup$ – John B Apr 26 '16 at 1:00
  • $\begingroup$ I don't understand, can you show me your steps? $\endgroup$ – hlapointe Apr 26 '16 at 1:01
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    $\begingroup$ I leave it to you. Sometimes you really need to make an effort, you do learn more. $\endgroup$ – John B Apr 26 '16 at 1:03
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The ODE $u'=1+u^2$ has as its solutions the functions $u(x)=\tan(x-c)$, as you have found out yourself. Since we want $u(0)=y'(0)=0$ we necessarily have $c=0$, so that we finally obtain $$y(x)=y(0)+\int_0^x u(t)\>dt=\int_0^x\tan t\>dt=\log(\sec x)\qquad\left(-{\pi\over2}<x<{\pi\over2}\right)\ .$$

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