# We have the following differential equation uu''=1+(u')^2 i found that the general

We have the following differential equation
$\cup {}^{″}=1+{\left({u}^{\prime }\right)}^{2}$
i found that the general solution of this equation is
$u=d\text{cosh}\left(\left(x-\frac{b}{d}\right)$
where b and d are constats
Please how we found this general solution?
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Alicia Washington
You can separate that equation as
$\frac{2{u}^{\prime }u{}^{″}}{1+{u}^{2}}=2\frac{{u}^{\prime }}{u}$
where both sides are complete differentials which integrate to
$\mathrm{ln}\left(1+{u}^{\prime 2}\right)=\mathrm{ln}\left({u}^{2}\right)+c⇒1+{u}^{\prime 2}=C{u}^{2}$
Can you continue?
###### Not exactly what you’re looking for?
Ourst1977
$\cup {}^{″}=1+{\left({u}^{\prime }\right)}^{2}$
Substitute $p={u}^{\prime }$
$u\frac{dp}{dx}=1+{p}^{2}$
$u\frac{dp}{du}\frac{du}{dx}=1+{p}^{2}$
$u\frac{dp}{du}p=1+{p}^{2}$
Now it's separable
$\int \frac{pdp}{1+{p}^{2}}=\int \frac{du}{u}$
It should be easy to integrate now..
Edit
${p}^{2}+1=K{u}^{2}⇒\int \frac{du}{\sqrt{K{u}^{2}-1}}=±x+{K}_{2}$
$\frac{ar\text{cosh}\left(\sqrt{K}u\right\}\left\{\sqrt{K}\right\}=x+{K}_{2}}{}$
Taking $\text{cosh}$ on both side
$\sqrt{K}u=\text{cosh}\left(\sqrt{K}\left(x+{K}_{2}\right)\right)$
$u=\frac{1}{\sqrt{K}}\text{cosh}\left(\sqrt{K}x+{K}_{2}\right)$
Which is close to your formula
$u=d\text{cosh}\left(\frac{x-b}{d}\right)⇒d=\frac{1}{\sqrt{K}}$ and $-\frac{b}{d}={K}_{2}$