MAT244--2018F > Quiz-5

Q5 TUT 0701

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Victor Ivrii:
Use the method of variation of parameters (without reducing the order) to determine the general solution of the given differential equation:
$$y''' - y' = \tanh(t).$$

Tzu-Ching Yen:
characteristic equation is
$$r^3 - r = 0\\
r(r+1)(r-1) = 0\\
r = 0, 1, \text{  or   } -1$$
The solution to homogeneous equations are $c_1e^t, c_2e^{-t}, c_3$
Let $y_1 = e^t, y_2 = e^{-t}, y_3 = 1$
$$W(y_1, y_2, y_3) = y_1'y_2'' - y_2'y_1'' = 1 + 1 = 2$$
Let $W_i$ be the determinant of wronskian with ith column substituted with $(0, 0, 1)$
$$W_1(t) = -e^{-t}\\
$$W_2(t) = e^{t}\\
$$W_3(t) = -2$$
given that $g(t) = \tanh(t)$
$$Y_1 = y_1 \int \dfrac{W_1(s)  g(s)}{W(s)}ds = \dfrac{-e^t}{2}(\int \tanh(s)e^{-t}ds) = -\dfrac{1}{2} + e^t\arctan(e^{-x})\\
Y_2 = y_2 \int \dfrac{W_2(s)  g(s)}{W(s)}ds = \dfrac{e^{-t}}{2}(\int \tanh(s)e^{t}ds) = \dfrac{1}{2} - e^{-t}\arctan(e^{x})\\
Y_3 = \int -tanh(s) ds =  -\ln(\cosh(x))$$
The general solution is therefore
$$y = c_1e^{-t} + c_2e^t + c_3 + e^t\arctan(e^{-x}) - e^{-t}\arctan(e^{x}) - \ln(\cosh(x))$$

Victor Ivrii:
Some obvious misprints corrected:

$$Y_1 = y_1 \int \dfrac{W_1(s)  g(s)}{W(s)}ds = \dfrac{-e^t}{2}(\int \tanh(s)e^{-s}ds) = -\dfrac{1}{2} + e^t\arctan(e^{-t})\\
Y_2 = y_2 \int \dfrac{W_2(s)  g(s)}{W(s)}ds = \dfrac{e^{-t}}{2}(\int \tanh(s)e^{t}ds) = \dfrac{1}{2} - e^{-t}\arctan(e^{t})\\
Y_3 = \int -\tanh(s) ds =  -\ln(\cosh(x))$$
The general solution is therefore
$$y = c_1e^{-t} + c_2e^t + c_3 + e^t\arctan(e^{-t}) - e^{-t}\arctan(e^{t}) - \ln(\cosh(t))$$

On the other hand $\arctan(e^{-t})=\frac{\pi}{2}-\arctan (e^t)$ and accommodating $\frac{\pi}{2}$ into constant we get
$$y = c_1e^{-t} + c_2e^t + c_3 - 2\cosh(t)\arctan(e^{-t}) - \ln(\cosh(t))$$

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