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I'am trying to answer a question from Michael D.Greenberg's Advanced Engineering Mathematics concerning a PDE.[chapter 1:introduction to modeling Ex1.2 Q4]

Verify that $u(x,t) = (Ax + B)(Ct + D) + (E \sin Kx + F \cos Kx)(G \sin Kct + H \cos Kct)$ is a solution of the one dimensional wave equation, $$c^2 \frac{\partial^2u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}$$ for any constants $A,B....H,K.$

So far i've tried differentiating $u(x,t) = (Ax + B)(Ct + D) + (E \sin Kx + F \cos Kx)(G \sin Kct + H \cos Kct)$ partially and i get

$\frac{\partial u}{\partial x} = (A + B) + (E \cos Kx - F \sin Kx)$ and $\frac{\partial u}{\partial t} = (C + D) + (G \cos Kc - H \sin Kc)$

now i don't know how to take it from here.first of all i'am unaware if i've partially differentiated it correctly.

Please Help

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You didn't get your partial derivative right. –  Paul Dec 23 '11 at 7:37

2 Answers 2

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It seems to me you're having trouble with the rules of differentiation, and the key here is to use the rules of differentiation to break the problem up. Three rules to start with are that $\frac{\partial}{\partial x}(f(x,t)+g(x,t)) = \frac{\partial}{\partial x}f(x,t) + \frac{\partial}{\partial x}g(x,t)$ (you can "break up" sums) and $\frac{\partial}{\partial x}f(x,t)h(t) = h(t)\frac{\partial}{\partial x}f(x,t)$ (functions of variables other than the one were differentiating with respect to can be "moved out") for any functions $f(x,t),g(x,t),h(t)$, and that the derivative of a constant is $0$. Applying the first rule, we get $$\frac{\partial}{\partial x}((Ax + B)(Ct + D)) + \frac{\partial}{\partial x}((E \sin Kx + F \cos Kx)(G \sin Kct + H \cos Kct))$$ the second gives us $$(Ct + D)\frac{\partial}{\partial x}(Ax + B) + (G \sin Kct + H \cos Kct)\frac{\partial}{\partial x}(E \sin Kx + F \cos Kx)$$ using the first again gives $$(Ct + D)(\frac{\partial}{\partial x}(Ax) + \frac{\partial}{\partial x}B) + (G \sin Kct + H \cos Kct)(\frac{\partial}{\partial x}(E \sin Kx) + \frac{\partial}{\partial x}(F \cos Kx))$$ which can be simplified using the second and third (remember, a constant can be considered a function of any variable we want) $$(Ct + D)A\frac{\partial}{\partial x}x + (G \sin Kct + H \cos Kct)(E\frac{\partial}{\partial x}(\sin Kx) + F\frac{\partial}{\partial x}(\cos Kx))$$ and further so by the fact that $\frac{\partial}{\partial x}x = 1$ to get $$(Ct + D)A + (G \sin Kct + H \cos Kct)(E\frac{\partial}{\partial x}(\sin Kx) + F\frac{\partial}{\partial x}(\cos Kx)).$$ The last thing we want to do is find $\frac{\partial}{\partial x}(\sin Kx)$ and $\frac{\partial}{\partial x}(\cos Kx)$, which requires what's called the chain rule. It is that for a function $f(x,t) = v(w(x,t))$, $\frac{\partial v}{\partial x} = \frac{\partial v}{\partial w}\frac{\partial w}{\partial x}$. In the case of $\sin Kx$, this means I can use $w(x) = Kx$ to find the derivative as follows: $\frac{\partial}{\partial x}(\sin Kx) = \frac{\partial}{\partial w}(\sin w) \frac{\partial w}{\partial x} = \cos w \times K = K\cos Kx$. This should help you find all the necessary derivatives to verify the equation. Keep in mind that even if a function doesn't look the same way the functions I used to explain these rules do, or if the variables are different in name or number, the rules still apply. Good luck!

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Thank you very much for taking so much of your time to explain the rules of differentiation.I prefer reading this rather than going back to my text. –  alok Dec 23 '11 at 8:33
    
You're welcome. If you have any more questions feel free to ask me. –  Alex Becker Dec 23 '11 at 8:35

Taking the partial derivative of $u$, we get: $$\frac{\partial u}{\partial x} = A(Ct+D)+K(E \cos Kx - F \sin Kx)(G \sin Kct + H \cos Kct),$$ $$\frac{\partial u}{\partial t} = C(Ax+B)+Kc(E \sin Kx + F \cos Kx)(G \cos Kct -H \sin Kct).$$ Taking the partial derivative one more time, we get $$\frac{\partial^2 u}{\partial x^2} = -K^2(E \sin Kx + F \cos Kx)(G \sin Kct + H \cos Kct),$$ $$\frac{\partial^2 u}{\partial t^2} = -K^2c^2(E \sin Kx + F \cos Kx)(G \sin Kct + H \cos Kct). $$ From this, you can see that $u$ satisfies $$c^2 \frac{\partial^2u}{\partial x^2} = \frac{\partial^2 u}{\partial t^2}.$$

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great,it is clear everything –  dato datuashvili Dec 23 '11 at 7:46

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