we have![\frac{\partial ^{2}u}{\partial t^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial t^{2}}](/tpl/images/0108/9221/b4868.png)
![=\frac{\partial }{\partial t}[\frac{\partial[f(x-at)+g(x+at)] }{\partial t}]](/tpl/images/0108/9221/8a2e0.png)
![\frac{\partial }{\partial t}[-a\cdot f'(x-at)+a\cdot g'(x+at)]](/tpl/images/0108/9221/574ca.png)
![=a^{2}[f''(x-at)+g''(x+at)].............(i)](/tpl/images/0108/9221/a9543.png)
![\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial x^{2}}](/tpl/images/0108/9221/8c3fd.png)
![=\frac{\partial }{\partial x}[\frac{\partial[f(x-at)+g(x+at)] }{\partial x}]](/tpl/images/0108/9221/b190f.png)
![=\frac{\partial }{\partial x}[f'(x-at)+g'(x+at)]](/tpl/images/0108/9221/62597.png)
show that the differential equation (on the left) is a solution of the function (on the right)
...
Mathematics, 19.07.2019 19:20 erees9653
show that the differential equation (on the left) is a solution of the function (on the right)
d^2u/dt^2 = a^2 * (d^2u/dx^2) u(x, t) = f(x-at) + g(x+at)
Answers: 2
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