I have a PDE in the form:
$$ \frac{\partial^2 y}{\partial t^2} + \frac{\partial^4 y}{\partial x^4} = 0 $$ with boundary conditions
$$ y(0,t) = 0 = y(1,t) $$ $$ \frac{\partial y}{\partial x}(0,t) = 0 = \frac{\partial^2 y}{\partial x^2}(1,t) $$
Initial conditions: $$ y(x,0) = 0.1 \sin \pi x, \quad \quad \frac{\partial y}{\partial t}(x,0) = 0 $$
I tried substituting
$$ y_1 = \frac{\partial y}{\partial t} \quad \quad y_2 = \frac{\partial^2 y}{\partial x^2} $$
$$ \frac{\partial y_1}{\partial t} = \frac{\partial^2 y}{\partial t^2} = -\frac{\partial ^4 y}{\partial x^4} = -\frac{\partial^2}{\partial x^2}(\frac{\partial^2 y}{\partial x^2}) = -\frac{\partial^2 y_2}{\partial x^2} $$
$$ \frac{\partial y_2}{\partial t} = \frac{\partial}{\partial t}(\frac{\partial^2 y}{\partial x^2}) = \frac{\partial^2}{\partial x^2}(\frac{\partial y}{\partial t}) = \frac{\partial^2 y_1}{\partial x^2} $$
so,
$$ \frac{\partial y_1}{\partial t} = -\frac{\partial^2 y_2}{\partial x^2}$$ $$ \frac{\partial y_2}{\partial t} = \frac{\partial^2 y_1}{\partial x^2}$$
Is this correct? How do I go from here?