Question

To transform + = — into a linear differential equation with constant coefficients, di2 dx x the required substitution is (a) x = sin t (b) x=t4+1 (c) x = log t (d) x = el 

Answer 1

$\displaystyle (a)\\ x = \sin{t} \\ \frac{\mathrm{d}x}{\mathrm{d}t} = \cos{t} \\ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\sin{t} \\ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = -x,\,\, \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + x = 0 \\ (b) \\ x = t^4 + 1 \\ \frac{\mathrm{d}x}{\mathrm{d}t} = 4t^3\\ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = 12t^2\\ \left(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right)^2 = 144t^4\\ \left(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right)^2 = 144(x - 1)\\ \left(\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right)^2 - 144(x - 1) = 0\\ (c) \\ x = \log{t} \\ \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{t} \\ \frac{\mathrm{d}x}{\mathrm{d}t} = e^{-x} \\ \frac{\mathrm{d}x}{\mathrm{d}t} - e^{-x} = 0\\ (d) \\ x = e^t \\ \frac{\mathrm{d}x}{\mathrm{d}t} = e^{t} = x \\ \frac{\mathrm{d}x}{\mathrm{d}t} - x = 0$