Question

Show that the function i) u(x, t) =A(x+ct) ^3 is a solution of the one-dimensional wave equation. ii) u(x, t) = {(e)^-(mu ×t)} sinx is a solution of the one-dimensional heat equation.

Answer 1

Answer on Question #70088 - Math - Differential Equations

Question

Show that the function

$i)$ $u(x,t) = A(x + ct)^3$ is a solution of the one-dimensional wave equation.

$ii)$ $u(x,t) = (e)^{-\mu t}\sin x$ is a solution of the one-dimensional heat equation.

Solution:

i)

$$u = (x_1 t) = A(x + ct)^3$$

The one-dimensional wave equation

$$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial x^2}$$ $$\frac{\partial u}{\partial t} = 3cA(x + ct)^2$$ $$\frac{\partial^2 u}{\partial t^2} = 6c^2A(x + ct)$$ $$\frac{\partial u}{\partial x} = 3A(x + ct)^2$$ $$\frac{\partial^2 u}{\partial x^2} = 6A(x + ct)$$

So

$$6c^2A(x + ct) = 6c^2A(x + ct)$$

ii)

$$u(x,t) = (e)^{-\mu t}\sin x$$

The one-dimensional heat equation

$$\frac{\partial u}{\partial t} = \mu \frac{\partial^2 u}{\partial x^2}$$ $$\frac{\partial u}{\partial t} = -\mu e^{-\mu t}\sin x$$ $$\frac{\partial u}{\partial x} = e^{-\mu t}\cos x$$ $$\frac{\partial^2 u}{\partial x^2} = -e^{-\mu t}\sin x$$

So

$$-\mu e^{-\mu t}\sin x = -\mu e^{-\mu t}\sin x$$

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