Ouchimdead's Question from Math Help Forum

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In summary, to show that \(y(x,t)=f(x-ct)+g(x+ct)\) is a solution to the given 2nd order differential equation, we can find the second derivatives of \(y(x,t)\) with respect to \(x\) and \(t\) and see that they satisfy the equation. This can be done by using substitution and following the steps outlined in the conversation.
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Sudharaka
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Title: how do i show something is a solution without solving it? 2nd order DE

the wave equation

δ2y/δx2 = (1/c2)(δ2y/δt2)

show y(x,t) = f(x-ct)+g(x+ct) is a solution explicitly


please show me how, don't just say "substitution"

Hi ouchimdead, :)

The only thing that you have to do here is to find the second derivatives of \(y(x,t)\) with respect to \(x\) and \(t\). Then it could be easily seen that they satisfy the given differential equation. :)

\[y(x,t) = f(x-ct)+g(x+ct)\]

\[\Rightarrow\frac{\partial}{\partial x}y(x,t) = \frac{\partial}{\partial x}f(x-ct)+\frac{\partial}{\partial x}g(x+ct)\mbox{ and }\frac{\partial}{\partial t}y(x,t) = -c\frac{\partial}{\partial t}f(x-ct)+c\frac{\partial}{\partial t}g(x+ct)\]

\[\Rightarrow\frac{\partial^{2}}{\partial x^2}y(x,t) = \frac{\partial^{2}}{\partial x^2}f(x-ct)+\frac{\partial^2}{\partial x^2}g(x+ct)\mbox{ and }\frac{\partial^2}{\partial t^2}y(x,t) = c^2\frac{\partial^2}{\partial t^2}f(x-ct)+c^2\frac{\partial^2}{\partial t^2}g(x+ct)\]

\[\therefore \frac{\partial^{2}}{\partial x^2}y(x,t)=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}y(x,t)\]

Kind Regards,
Sudharaka.
 

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