We have the following differential equation∪″=1+(u′)2i found that the general solution of this equation isu=dcosh((x−bd)where...

jazzcutie0h

Answered question

2021-11-19

We have the following differential equation $\cup {}^{\u2033}=1+{\left({u}^{\prime}\right)}^{2}$ i found that the general solution of this equation is $u=d\text{cosh}((x-\frac{b}{d})$ where b and d are constats Please how we found this general solution?

Answer & Explanation

Alicia Washington

Beginner2021-11-20Added 23 answers

You can separate that equation as $\frac{2{u}^{\prime}u{}^{\u2033}}{1+{u}^{2}}=2\frac{{u}^{\prime}}{u}$ where both sides are complete differentials which integrate to $\mathrm{ln}(1+{u}^{\prime 2})=\mathrm{ln}\left({u}^{2}\right)+c\Rightarrow 1+{u}^{\prime 2}=C{u}^{2}$ Can you continue?

Ourst1977

Beginner2021-11-21Added 21 answers

$\cup {}^{\u2033}=1+{\left({u}^{\prime}\right)}^{2}$ 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 $p}^{2}+1=K{u}^{2}\Rightarrow \int \frac{du}{\sqrt{K{u}^{2}-1}}=\pm x+{K}_{2$ $\frac{ar\text{cosh}\left(\sqrt{K}u\right\}\left\{\sqrt{K}\right\}=x+{K}_{2}}{}$ Taking $\text{cosh}$ on both side $\sqrt{K}u=\text{cosh}\left(\sqrt{K}(x+{K}_{2})\right)$ $u=\frac{1}{\sqrt{K}}\text{cosh}(\sqrt{K}x+{K}_{2})$ Which is close to your formula $u=d\text{cosh}\left(\frac{x-b}{d}\right)\Rightarrow d=\frac{1}{\sqrt{K}}$ and $-\frac{b}{d}={K}_{2}$