Kindiak Gauntlet (Differential Equations)

Warning. All questions below are non-trivial.

Section A (48 marks)

Each question in this section is worth 8 marks.

Question 1. Solve the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{1 + 2y^2}{\sqrt{9 - 4x^2}}.$

(Click for Solution)

Solution. By the method of separable variables,

$\begin{aligned} \frac{1}{1 +2y^2} \cdot \frac{\mathrm dy}{\mathrm dx} &= \frac 1{\sqrt{9-4x^2}} \\ \frac 12 \cdot \frac{1}{\left( \frac 1{\sqrt 2} \right)^2 +y^2} \cdot \frac{\mathrm dy}{\mathrm dx} &= \frac 1 2 \cdot \frac 1{\sqrt{ \left( \frac 32 \right)^2 - x^2}} \\ \int \frac{1}{\left( \frac 1{\sqrt 2} \right)^2 +y^2}\, \mathrm dy &= \int \frac 1{\sqrt{ \left( \frac 32 \right)^2 - x^2}}\, \mathrm dx \\ \frac 1{\frac 1{\sqrt 2}} \tan^{-1} \left( \frac{y}{\frac 1{\sqrt 2}} \right) &= \sin^{-1}\left( \frac{x}{\frac 32} \right) + C \\ \sqrt{2} \tan^{-1}(y \sqrt 2) &= \sin^{-1}\left( \frac {2x}3\right) + C. \end{aligned}$

Question 2. Solve the differential equation

$\displaystyle x \left( \frac{\mathrm dy}{\mathrm dx} - \cos(x) \right) + 2y = 0.$

(Click for Solution)

Solution. By algebraic manipulation,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} + \frac{2y}{x} = \cos(x).$

Compute the integrating factor

$\displaystyle I(x) = e^{\int \frac 2x\, \mathrm dx} = e^{2 \ln|x|} = |x|^2 = x^2.$

Thus, the general solution is given by

$\displaystyle \begin{aligned} x^2 y &= \int x^2 \cos(x)\, \mathrm dx \\ &= x^2 \sin(x) - \int 2x \sin(x)\, \mathrm dx \\ &= x^2 \sin(x) + 2 \int x\cdot(-\sin(x))\, \mathrm dx \\ &= x^2 \sin(x) + 2\left(x \cos(x) - \int \cos(x)\, \mathrm dx \right) \\ &= x^2 \sin(x) + 2x \cos(x) - 2 \sin(x) + C. \end{aligned}$

where we integrated by parts twice on the right-hand side.

Question 3. Use the substitution $u = xy$ to solve the differential equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2} - \left( 5 - \frac 2x \right) \frac{\mathrm dy}{\mathrm dx} - \left( \frac 5x - 6 \right) y = 0.$

(Click for Solution)

Solution. Writing $y = u/x$ , use the quotient rule to deduce

$\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &= \frac{xu' - u}{x^2} = \frac{u'}{x} - \frac{u}{x^2},\\ \frac{\mathrm d^2y}{\mathrm dx^2} &= \frac{xu'' - u'}{x^2} - \frac{x^2u' - u \cdot 2x}{x^4} = \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3}.\end{aligned}$

Making the substitutions,

$\begin{aligned} \left( \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3} \right) - \left( 5 - \frac 2x \right) \left( \frac{u'}{x} - \frac{u}{x^2} \right) - \left( \frac 5x - 6 \right) \frac ux &= 0 \\ \left( \frac{u''}{x} - \frac{2u' }{x^2} + \frac{2u}{x^3} \right) - \left( \frac{5u'}{x} - \frac{5u}{x^2} -\frac{2u'}{x^2} + \frac{2u}{x^3} \right) - \left( \frac {5u}{x^2} - \frac{6u}{x} \right) &= 0 \\ \frac{u''}{x} - \frac{5u'}{x} + \frac{6u}{x} &= 0 \\ u'' - 5u' + 6u &= 0. \end{aligned}$

Since the auxiliary equation $m^2 - 5m + 6 = 0$ has real and distinct roots $m = 2,3$ , the general solution is given by

$\displaystyle xy = u = C_1 e^{2x} + C_2 e^{3x}.$

Question 4. Evaluate $\displaystyle \mathcal L\{(7+8t)^2 * (4 \sin 3t \cos 3t)\}$ .

(Click for Solution)

Solution. First carry out some simplifications:

$\begin{aligned} (7+8t)^2 &= 49 + 112t + 64t^2, \\ 4\sin 3t \cos 3t &= 2 \sin 6t.\end{aligned}$

Then taking Laplace transforms,

$\begin{aligned} \mathcal L\{(7+8t)^2 * (4 \sin 3t \cos 3t)\} &= \mathcal L\{(7+8t)^2 \} \cdot \mathcal L\{ 4 \sin 3t \cos 3t \} \\ &= \mathcal L\{ 49 + 112t + 64t^2 \} \cdot \mathcal L\{ 2 \sin 6t \} \\ &= \left( 49 \cdot \frac 1s + 112 \cdot \frac{1}{s^2} + 64 \cdot \frac{2}{s^3} \right) \cdot \left( 2 \cdot \frac{6}{s^2 + 6^2} \right) \\ &= \frac{49s^2 + 112s + 128}{s^3} \cdot \frac{12}{s^2 + 36} \\ &= \frac{12(49s^2 + 112s + 128)}{s^3 ( s^2 + 36 )}. \end{aligned}$

Question 5. Evaluate $\displaystyle \mathcal L^{-1} \left\{ \frac{1}{s^3+1} \right\}$ .

(Click for Solution)

Solution. We first note that $(-1)^3 + 1 = 0$ , so that by the factor theorem, $s+1$ is a factor of $s^3+1$ . Hence, we factorise $s^3+1 = (s+1)(s^2 - s + 1)$ . This suggests us to employ the partial fraction decomposition

$\displaystyle \frac 1{s^3 + 1} = A \cdot \frac{1}{s+1} + B \cdot \frac{s-1/2}{(s-1/2)^2+ (\sqrt{3}/2)^2} + C \cdot \frac{1}{(s-1/2)^2+ (\sqrt{3}/2)^2},$

where we will obtain $A,B,C$ later and completed the square in the last term. We recall the following well-known inverse Laplace transforms:

$\displaystyle \begin{aligned} \mathcal L^{-1} \left\{ \frac{1}{s+1} \right\} &= e^{-(-1)t} = e^t, \\ \mathcal L^{-1} \left\{ \frac{s-1/2}{(s-1/2)^2+ (\sqrt{3}/2)^2} \right\} &= e^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right), \\ \mathcal L^{-1} \left\{ \frac{1}{(s-1/2)^2+ (\sqrt{3}/2)^2} \right\} &= \frac 2{\sqrt 3} e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right). \end{aligned}$

Since taking inverse Laplace transforms is linear,

$\displaystyle \mathcal L^{-1}\left\{ \frac{1}{s^3+1} \right\} = Ae^t + Be^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right) + \frac {2C}{\sqrt 3} e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right).$

All that remains is for us to obtain $A, B, C$ . To that end, clear denominators by multiplying $x^3+1$ on all sides:

$\displaystyle \begin{aligned} 1 &= A(s^2 - s + 1) + B(s-1/2)(s+1) + C(s+1) \\ &= (A + B)s^2 + (-A +B/2 + C)s + (C-B/2). \end{aligned}$

Comparing coefficients,

$\left\{ \begin{aligned}A + B &= 0, \\ -A + \frac 12 B + C &= 0, \\ -\frac 12 B + C &= 1.\end{aligned} \right.$

Solving simultaneous linear equations, $A=1/2,B = -1/2,C = 3/4$ . Substituting and simplifying,

$\displaystyle \mathcal L^{-1}\left\{ \frac{1}{s^3+1} \right\} = \frac 12 e^t - \frac 12 e^{t/2} \cos\left( \frac{t\sqrt 3}{2} \right) + \frac{\sqrt 3}2 e^{t/2} \sin\left( \frac{t\sqrt 3}{2} \right).$

Question 6. Evaluate $\displaystyle \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)}\left( \sin \frac 23 x \right).$

(Click for Solution)

Solution. Apply partial fractions again (left as an exercise) to obtain

$\displaystyle \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)} = \frac{2}{4 + 9 \mathcal D^2} - \frac{3}{16 + 25 \mathcal D^2}.$

Therefore,

$\begin{aligned} \frac {20 + 23 \mathcal D^2}{(4 + 9 \mathcal D^2)(16 + 25 \mathcal D^2)} \left( \sin \frac 23 x \right) &= \left(\frac{2}{4 + 9 \mathcal D^2} - \frac{3}{16 + 25 \mathcal D^2}\right) \left( \sin \frac 23 x \right) \\ &= \frac 29 \cdot \frac{1}{\mathcal D^2 + \left(\frac 23\right) ^2}\left( \sin \frac 23 x \right) - 3 \cdot \frac{1}{16 + 25 \mathcal D^2}\left( \sin \frac 23 x \right) \\ &= -\frac 29 \cdot \frac{x \cos \left(\frac 23 x\right)}{2 \cdot \frac 23} - 3 \cdot \frac 1{16 + 25 \cdot \left(-\frac 49\right)} \cdot \sin \left(\frac 23 x \right) \\ &= -\frac 16 x \cos \left(\frac 23 x\right) - \frac{27}{44} \sin \left(\frac 23 x \right).\end{aligned}$

Section B (52 marks)

Each question in this section is worth 13 marks.

Question 1. Solve the differential equation

$\displaystyle 3y'' - 24y' + 48y = e^{5x} \cos(6x).$

(Click for Solution)

Solution. The auxiliary equation $3m^2 - 24m + 48 = 0$ has repeated real roots $m = 4$ . Therefore, the complementary function is given by

$y_{\mathrm C} = (C_1x + C_2)e^{4x}.$

For the particular integral,

$\begin{aligned} y_{\mathrm P} &= \frac{1}{3 \mathcal D^2 - 24 \mathcal D + 48}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot \frac{1}{\mathcal D^2 - 8 \mathcal D + 16}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{(\mathcal D + 5)^2 - 8 (\mathcal D + 5) + 16}(e^{5x} \cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{(\mathcal D^2 - 10 \mathcal D + 25) - (8\mathcal D + 40) + 16}(\cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{\mathcal D^2 - 18 \mathcal D + 1}(\cos(6x)) \\ &= \frac 13 \cdot e^{5x} \cdot \frac{1}{-6^2 - 18 \mathcal D + 1}(\cos(6x)) \\ &= -\frac 13 \cdot e^{5x} \cdot \frac{1}{18 \mathcal D + 35}(\cos(6x)) \\ &= -\frac 13 \cdot e^{5x} \cdot \frac{18 \mathcal D - 35}{324 \mathcal D^2 - 1225}(\cos(6x)) \\ &= \frac 1{38\, 667} \cdot e^{5x} \cdot (18 \mathcal D(\cos(6x)) - 35\cos(6x)) \\ &= \frac 1{38\, 667} \cdot e^{5x} \cdot (-108 \sin(6x) - 35\cos(6x)) \\ &= -\frac 1{38\, 667} \cdot e^{5x} \cdot (108 \sin(6x) + 35\cos(6x)). \end{aligned}$

Therefore, the general solution is given by

$y = -\frac 1{38\, 667} e^{5x} (108 \sin(6x) + 35\cos(6x)) + (C_1x + C_2)e^{4x}.$

Question 2. Given the function $f$ defined by

$\displaystyle f(t) = \begin{cases}2, & t < 1, \\ 3, & 1 \leq t < 2, \\ 5, & 2 \leq t < 3, \\ 7 , & t \geq 3,\end{cases}$

evaluate $\displaystyle \mathcal L\left\{ \sqrt{f(\sqrt{t})} \cdot U(t - 5)\right\}$ .

(Click for Solution)

Solution. We first observe that

$\displaystyle \sqrt{ f(\sqrt{t}) } = \begin{cases}\sqrt 2, & \sqrt{t} < 1, \\ \sqrt 3, & 1 \leq \sqrt{t} < 2, \\ \sqrt 5, & 2 \leq \sqrt{t} < 3, \\ \sqrt 7 , & \sqrt{t} \geq 3\end{cases} = \begin{cases}\sqrt 2, & t < 1, \\ \sqrt 3, & 1 \leq t < 4, \\ \sqrt 5, & 4 \leq t < 9, \\ \sqrt 7 , & t \geq 9.\end{cases}$

Therefore,

$\displaystyle \sqrt{ f(\sqrt{t}) } \cdot U(t-5) = 0 + \sqrt{5} \cdot U(t-5) + (\sqrt{7} - \sqrt{5}) \cdot U(t-9).$

Taking Laplace transforms,

$\displaystyle \mathcal L\left\{ \sqrt{ f(\sqrt{t}) } \cdot U(t-5) \right\} = \frac 1s \left( \sqrt 5 e^{-5s} + (\sqrt 7 - \sqrt 5) e^{-9s} \right).$

Question 3. Use Laplace transforms to solve the initial value problem

$\displaystyle \left\{ \begin{aligned} y'' - 2y' + 3y &= t^2 \delta(t - 4), \\ y(0) &= 5, \\ y'(0) &= 6.\end{aligned} \right.$

(Click for Solution)

Solution. Taking Laplace transforms on both sides,

$\mathcal L\{y''\} - 2 \mathcal L\{y'\} + 3 \mathcal L\{y\} = \mathcal L\{t^2 \delta(t-4)\}.$

To evaluate the left-hand side,

$\begin{aligned} \mathcal L\{y''\} &= s^2 Y(s) - sy(0) - y'(0) = s^2 Y(s) - 5s - 6, \\ \mathcal L\{y'\} &= s Y(s) - y(0) = s Y(s) - 5, \\ \mathcal L\{y\} &= Y(s). \end{aligned}$

Therefore, the left-hand side simplifies to

$\begin{aligned} & \mathcal L\{y''\} - 2 \mathcal L\{y'\} + 3 \mathcal L\{y\} \\ &= (s^2 Y(s) - 5s - 6) - 2(s Y(s) - 5) + 3Y(s) \\ &= (s^2 - 2s + 3)Y(s) - 5s +4. \end{aligned}$

To evaluate the right-hand side,

$\begin{aligned} \mathcal L\{t^2 \delta(t-4) \} &= -\frac{\mathrm d}{\mathrm ds} \mathcal L\{t \delta(t-4) \} \\ &= -\frac{\mathrm d}{\mathrm ds} \left( -\frac{\mathrm d}{\mathrm ds} \mathcal L\{ \delta(t-4) \} \right) \\ &= -\frac{\mathrm d}{\mathrm ds} \left( -\frac{\mathrm d}{\mathrm ds} (e^{-4s}) \right) = 16e^{-4s}. \end{aligned}$

Putting it all together,

$(s^2 - 2s + 3)Y(s) - 5s +4 = 16e^{-4s}.$

By algebraic manipulation,

$\begin{aligned} Y(s) &= \frac{16 e^{-4s}}{s^2 - 2s + 3} + \frac{5s-4}{s^2 - 2s + 3} \\ y(t) &= 16 \cdot \mathcal L^{-1}\left\{ e^{-4s} \cdot \frac{1}{s^2 - 2s + 3} \right\} + \mathcal L^{-1} \left\{ \frac{5s-4}{s^2 - 2s + 3} \right\}. \end{aligned}$

It remains to (painfully) compute each of the following inverse Laplace transforms, by observing that $s^2 - 2s + 3 = (s-1)^2 + 2$ .

Define $G(s-1) = (5s-4)/((s-1)^2 + 2)$ so that

$\displaystyle G(s) = \frac{5(s+1)-4}{s^2 + 2} = \frac{5s+1}{s^2 + 2} = 5 \cdot \frac s{s^2+2} + \frac 1{s^2+2}.$

By the shift theorems,

$\begin{aligned}\mathcal L^{-1} \left\{ \frac{5s-4}{s^2 - 2s + 3} \right\} &= \mathcal L^{-1}\{ G(s-1)\} \\ &= e^t g(t) \\ &= e^t \cdot \mathcal L^{-1}\left\{ 5 \cdot \frac s{s^2+2} + \frac 1{s^2+2} \right\} \\ &= e^t \cdot \left(5 \cos (t\sqrt 2) + \frac 1{\sqrt 2} \sin(t \sqrt 2) \right).\end{aligned}$

Similarly, defining $H(s) = 1/(s^2 + 2)$ and $H_1 := H(\cdot - 1)$ so that

$\displaystyle h_1(t) = \mathcal L^{-1}\{ H_1(s)\} = \mathcal L^{-1} \{H(s-1)\} = e^t \cdot h(t) = e^t \cdot \frac{1}{\sqrt 2} \sin(t \sqrt 2),$

we have

$\begin{aligned} \mathcal L^{-1}\left\{ e^{-4s} \cdot \frac{1}{s^2 - 2s + 3} \right\} &= \mathcal L^{-1} \{e^{-4s} H_1(s)\} \\ &= h_1(t-4) \cdot U(t-4) \\ &= \frac 1{\sqrt 2} e^{t-4} \sin((t-4)\sqrt 2) \cdot U(t-4). \end{aligned}$

Piecing them all together,

$\displaystyle y(t) = 8\sqrt 2 \cdot e^{t-4} \sin((t-4)\sqrt 2) \cdot U(t-4) + e^t \cdot \left(5 \cos (t\sqrt 2) + \frac 1{\sqrt 2} \sin(t \sqrt 2) \right).$

Question 4. Evaluate the Fourier series of the function $f$ defined by

$f(t) = \begin{cases} 1, & -1 \leq t \leq 0, \\ 2t, & 0 \leq t \leq 1, \end{cases}$

and periodic extension $f(t) = f(t+2)$ .

(Click for Solution)

Solution. We need to evaluate each of the Fourier coefficients $a_0, a_n, b_n$ with $T = 2$ . For $a_0$ ,

$\begin{aligned} a_0 = \int_{-1}^1 f(t)\, \mathrm dt &= \int_{-1}^0 1\, \mathrm dt + \int_0^1 2t\, \mathrm dt \\ &= (0 - (-1)) + (1^2 - 0^2) = 2. \end{aligned}$

For $a_n$ , we need to integrate by parts:

$\begin{aligned} a_n &= \int_{-1}^1 f(t) \cos(n \pi t) \, \mathrm dt \\ &= \int_{-1}^0 \cos(n \pi t)\, \mathrm dt + \int_0^1 2t \cos(n \pi t)\, \mathrm dt \\ &= \left[ \frac{\sin(n \pi t)}{n \pi} \right]_{-1}^0 + \left[ \frac{2 t \sin(n \pi t)}{n \pi} \right]_0^1 + \left[ \frac{2\cos(n \pi t)}{n^2 \pi^2} \right]_0^1 \\ &= 0 + 0 + \frac 2{n^2 \pi^2} ((-1)^n - 1) \\ &= \frac {2 ((-1)^n - 1) }{n^2 \pi^2} . \end{aligned}$

Similarly for $b_n$ , we need to integrate by parts:

$\begin{aligned} b_n &= \int_{-1}^1 f(t) \sin(n \pi t) \, \mathrm dt \\ &= \int_{-1}^0 \sin(n \pi t)\, \mathrm dt + \int_0^1 2t \sin(n \pi t)\, \mathrm dt \\ &= \left[ \frac{-\cos(n \pi t)}{n \pi} \right]_{-1}^0 + \left[ \frac{-2 t \cos(n \pi t)}{n \pi} \right]_0^1 + \left[ \frac{2\sin(n \pi t)}{n^2 \pi^2} \right]_0^1 \\ &= -\frac 1{n\pi} (1 - (-1)^n) - \frac 2{n \pi} (-1)^n \\ &= -\frac 1{n\pi} ( (-1)^n + 1). \end{aligned}$

Therefore, the Fourier series of $f$ is given by

$\displaystyle f(t) = 1 + \sum_{n=1}^\infty \left( \frac {2 ((-1)^n - 1) }{n^2 \pi^2} \cos(n \pi t) -\frac 1{n\pi} ( (-1)^n + 1) \sin(n \pi t) \right).$

—Joel Kindiak, 19 Aug 25, 1613H

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