We have the following differential equation∪″=1+(u′)2i found that the general solution of this equation isu=dcosh((x−bd)where b and d are constatsPlease how we found this general solution?

Name: We have the following differential equation∪″=1+(u′)2i found that the general solution of this equation isu=dcosh((x−bd)where b and d are constatsPlease how we found this general solution?
Uploaded: 2022-02-23T17:22:57+00:00
Description: We have the following differential equation∪″=1+(u′)2i found that the general solution of this equation isu=dcosh((x−bd)where b and d are constatsPlease how we found this general solution?

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Alicia Washington · Accepted Answer

You can separate that equation as2u′u″1+u2=2u′uwhere both sides are complete differentials which integrate toln⁡(1+u′2)=ln⁡(u2)+c⇒1+u′2=Cu2Can you continue?

Ourst1977 · Accepted Answer

∪″=1+(u′)2Substitute p=u′udpdx=1+p2udpdududx=1+p2udpdup=1+p2Now it&#039;s separable∫pdp1+p2=∫duuIt should be easy to integrate now..Editp2+1=Ku2⇒∫duKu2−1=±x+K2arcosh(Ku}{K}=x+K2Taking cosh on both sideKu=cosh(K(x+K2))u=1Kcosh(Kx+K2)Which is close to your formulau=dcosh(x−bd)⇒d=1K and −bd=K2

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

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