Day: September 26, 2020

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202009260344 Exercise 2.5.1

Show that x=e^{2\theta}\sin\theta satisfies the equation x''-4x'+5x=0.

Solution.

(bottom-up)

Let x' denote \displaystyle{\frac{\mathrm{d}x}{\mathrm{d}\theta}}, x'' denote \displaystyle{\frac{\mathrm{d}^2x}{\mathrm{d}\theta^2}}.

\begin{aligned} x' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \sin\theta \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( \sin\theta \big) + \sin\theta \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = e^{2\theta} \cos\theta + 2e^{2\theta}\sin\theta \\ & = e^{2\theta}(\cos\theta + 2\sin\theta ) \\ \end{aligned}

\begin{aligned} x'' & = \frac{\mathrm{d}}{\mathrm{d}\theta} \big( x' \big) \\ & = e^{2\theta} \cdot \frac{\mathrm{d}}{\mathrm{d}\theta} (\cos\theta +2\sin\theta ) + (\cos\theta + 2\sin\theta )\cdot \frac{\mathrm{d}}{\mathrm{d}\theta} \big( e^{2\theta} \big) \\ & = (e^{2\theta})(-\sin\theta + 2\cos\theta ) + (\cos\theta + 2\sin\theta )(2e^{2\theta }) \\ & = e^{2\theta}(3\sin\theta + 4 \cos\theta ) \\ \end{aligned}

To know whether or not x=e^{2\theta}\sin\theta is a solution, I simply do substitution in the equation x''-4x'+5x=0.

\begin{aligned} \textrm{LHS} & = x''-4x'+5x \\ & = e^{2\theta}(3\sin\theta +4\cos\theta )-4\big[ e^{2\theta}(\cos\theta +2\sin\theta )\big] + 5\big(e^{2\theta}\sin\theta \big) \\ & = \dots \\ & = 0 \\ & = \textrm{RHS} \\ \end{aligned}

Revision.

(top-down)

We are given a second-order linear homogeneous ordinary differential equation (ODE):

x''-4x'+5x = 0

with some independent variable \theta and some dependent variable x(\theta )=e^{2\theta }\sin\theta, the coefficients of x'', x', and x being constants 1, -4, and 5.

Let the primed [ *]' be the function derived wrt. to \theta. Following the routine procedures,

\begin{aligned} x & = e^{r\theta} \\ x' & = re^{r\theta }\\ x'' & = r\cdot ( e^{r\theta })' + r' \cdot ( e^{r\theta }) \\ & = r^2e^{r\theta} \end{aligned}

rewrite it,

\begin{aligned} x''-4x'+5x & = 0 \\ r^2e^{r\theta} - 4re^{r\theta} + 5e^{r\theta} & = 0 \\ \dots \textrm{\quad excepting\quad} & x=e^{r\theta}=0\textrm{\quad \dots} \\ r^2 -4r+5 & = 0 \\ \end{aligned}

and we shall obtain the auxiliary equation (aka. the characteristic equation) on the very last line.

r=\displaystyle{\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(5)}}{2(1)}}=2\pm \textrm{i}.

Note.

If the roots of the auxiliary equation ar^2+br+c=0 are the complex numbers r_1=\alpha +\textrm{i}\beta and r_2=\alpha -\textrm{i}\beta, the general solution of ay''+by'+cy=0 is

y=e^{\alpha x}(c_1\cos\beta x+c_2\sin\beta x)

(please refer to J. Stewart’s Calculus, Second-Order Linear Differential Equations)

Now that r_1=2+\textrm{i} and r_2=2-\textrm{i} and c_1, c_2 are arbitrary constants, to this 2^{\textrm{nd}}-order ODE, the general solution is

x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta ).

Examination.

It is a good practice, no matter how much time one would allow oneself, to countercheck one’s solution…

Here it goes,

\begin{aligned} x & =e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (c_2e^{2\theta })(\sin\theta ) + (c_1e^{2\theta })(\cos\theta ) \\ \end{aligned}.

\begin{aligned} x' & = e^{2\theta}(-c_1\sin\theta +c_2\cos\theta ) + 2e^{2\theta}(c_1\cos\theta +c_2\sin\theta ) \\ & = (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) + (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \\ \end{aligned}

\begin{aligned} x''& = \big[ (2c_2e^{2\theta }-c_1e^{2\theta})(\sin\theta ) \big]' + \big[ (2c_1e^{2\theta} + c_2e^{2\theta} )(\cos\theta ) \big]' \\ & = \big[ (2c_2e^{2\theta}-c_1e^{2\theta })(\cos\theta )+(4c_2e^{2\theta }-2c_1e^{2\theta })(\sin\theta ) \big] \\ & \qquad\quad + \big[ (2c_1e^{2\theta }+c_2e^{2\theta })(-\sin\theta )+(4c_1e^{2\theta }+2c_2e^{2\theta })(\cos\theta ) \big] \\ & = (-4c_1e^{2\theta }+3c_2e^{2\theta })(\sin\theta ) + (3c_1e^{2\theta }+4c_2e^{2\theta })(\cos\theta ) \\ \end{aligned}

Then,

\begin{aligned} \textrm{LHS} & = x'' - 4x' + 5x \\ & = \big( (-4c_1e^{2\theta }+3c_2e^{2\theta })-4(2c_2e^{2\theta}-c_1e^{2\theta})+5c_2e^{2\theta}\big)(\sin\theta ) \\ & \qquad\quad + \big( (3c_1e^{2\theta}+4c_2e^{2\theta}) - 4(2c_1e^{2\theta}+c_2e^{2\theta})+5c_1e^{2\theta} \big)(\cos\theta ) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}

In conclusion, x=e^{2\theta}\sin\theta is a particular solution, the general solution being x=e^{2\theta}(c_1\cos\theta +c_2\sin\theta ).