- Thread starter
- #1
- Feb 5, 2012
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Title: how do i show something is a solution without solving it? 2nd order DE
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.
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.