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$y=\cosh\lambda\cosh-\lambda at$ is a solution of $\frac{\partial^2y}{\partial t^2}=a^2\frac{\partial^2y}{\partial x^2}$

Here we have to prove whether this is true or false. I get $$F'(x)=\lambda\sinh\lambda x\cosh-\lambda at$$ $$F''(x)=\lambda^2 y$$ $$F'(y)=-\lambda a\sin\lambda x\sinh-\lambda at$$ $$F''(y)=\lambda^2a^2y$$ Thus both sides are equal, but the answer is false – why?

Parcly Taxel
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KD.
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  • why don't you type that out in mathjax? – klirk Dec 02 '17 at 13:13
  • i am not knowing how to write in mathjax – KD. Dec 02 '17 at 13:39
  • No problem. It improves readibility a lot however, so more people will look at your question. Take a look: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – klirk Dec 02 '17 at 13:45

2 Answers2

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$$\frac{\partial^2 y}{\partial x^2} = \lambda^2 \cosh (\lambda x) \cosh ( a \lambda t) $$

$$\frac{\partial^2 y}{\partial t^2} = a^2 \lambda^2 \cosh (\lambda x) \cosh ( a \lambda t) = a^2 \frac{\partial^2 y}{\partial x^2}$$

Hence the claim holds.

actinidia
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  • yeah i also done same but actually answers is that ∂2y∂t2=a2∂2y∂x2 are not equal FALSE is answer – KD. Dec 02 '17 at 13:33
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$$\frac{\partial y}{\partial x} = \lambda\sinh (\lambda x) \cosh (-\lambda at)$$

$$\frac{\partial^2 y}{\partial x^2} =\lambda^2\cosh (\lambda x) \cosh (-\lambda at)$$

$$\frac{\partial y}{\partial t} = -\lambda a \sinh (- \lambda at) \cosh (\lambda x)$$

$$\frac{\partial^2 y}{\partial t^2} =\lambda^2 a^2 \cosh (- \lambda at) \cosh (\lambda x)$$

$$\frac{\partial^2 y}{\partial t^2} = a^2\frac{\partial^2 y}{\partial x^2}$$

Maadhav
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  • yeah i also done same but actually answers is that ∂2y∂t2=a2∂2y∂x2 are not equal FALSE is answer – KD. Dec 02 '17 at 13:33
  • @KD. Your answer key might be wrong – Maadhav Dec 02 '17 at 13:59
  • @KD. I fed the equations to wolfram and wolfram agrees too. https://www.wolframalpha.com/input/?i=differentiate+twice&rawformassumption=%7B%22F%22,+%22SecondDerivativeCalculator%22,+%22derivativefunction%22%7D+-%3E%22y%3Dcosh+(lambda+x)+cosh+(-a+lambda+t)%22&rawformassumption=%7B%22F%22,+%22SecondDerivativeCalculator%22,+%22derivativevariable%22%7D+-%3E%22x%22 – Maadhav Dec 02 '17 at 14:00
  • https://www.wolframalpha.com/input/?i=differentiate+twice&rawformassumption=%7B%22F%22,+%22SecondDerivativeCalculator%22,+%22derivativefunction%22%7D+-%3E%22y%3Dcosh+(lambda+x)+cosh+(-a+lambda+t)%22&rawformassumption=%7B%22F%22,+%22SecondDerivativeCalculator%22,+%22derivativevariable%22%7D+-%3E%22t%22 – Maadhav Dec 02 '17 at 14:00
  • @KD. Your answer is correct. LHS=RHS. – Maadhav Dec 02 '17 at 14:01