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I have to find the general solution for the differential equation $$y''(x) - m^2y(x) = 0$$I tried to solve this one by deriving the auxiliary equation as $M^2-m^2=0$ which gives $M = \pm m$ hence the general solutions is $c_1e^{mx} + c_2e^{-mx}$ but in the paper it's given the answer should be in the form of summation of hyperbolic functions,which is $c_1 \sinh mx + c_2 \cosh mx$

I haven't done anything much on hyperbolic,could anybody help me in this regard?

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$$c_1e^{mx} + c_2e^{-mx}$$

and

$$c_3\cosh\,mx + c_4\sinh\,mx$$

are two different ways of writing the same thing, since $\cosh\,x=\frac12(e^x+e^{-x})$ and $\sinh\,x=\frac12(e^x-e^{-x})$. You should be able to figure out how to express $c_1$ and $c_2$ in terms of $c_3$ and $c_4$, and vice-versa, to go back and forth between these two forms.

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    $\begingroup$ Does $\cosh\,mx=\frac12(e^{mx}+e^{-mx})$ and similarly for $\sinh\,mx$ too? $\endgroup$ – Quixotic May 6 '11 at 18:47
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    $\begingroup$ Yes.$\qquad\quad$ $\endgroup$ – J. M. is a poor mathematician May 6 '11 at 18:48
  • $\begingroup$ @J.M:I am trying but not getting there,I never really did anything over hyperbolic may be cold feet because of that,could you please give me some more hints how to convert between the forms quickly as this is a objective question and I am suppose to solve it fast. $\endgroup$ – Quixotic May 6 '11 at 18:56
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    $\begingroup$ Let $c_1=\frac{c_3+c_4}{2}$ and $c_2=\frac{c_3-c_4}{2}$... $\endgroup$ – J. M. is a poor mathematician May 6 '11 at 19:00
  • $\begingroup$ sorry,I could not understand :( $\endgroup$ – Quixotic May 6 '11 at 19:07

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