Problems

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Solution: \begin{align*} && x &= \frac{1}{t^2-1} \\ && t &= \sqrt{\frac{x+1}{x}}\\ \Rightarrow && \frac{\d x}{\d t} &= \frac{-2t}{(t^2-1)^2} \\ \Rightarrow && I &= \int \frac{1}{\sqrt{x(x+1)}} \d x \\ &&&= \int \frac{1}{\sqrt{\frac1{t^2-1} \frac{t^2}{t^2-1}}} \cdot \frac{-2 t}{(t^2-1)^2} \d t \\ &&&= \int \frac{t^2-1}{t} \frac{-2t}{(t^2-1)^2} \d t \\ &&&= -\int \frac{2}{t^2-1} \d t \\ &&&= - \frac{2}{2 \cdot 1} \ln \left | \frac{t-1}{t+1} \right| +C \\ &&&= \ln \left | \frac{t+1}{t-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{\frac{x+1}{x}}+1}{\sqrt{\frac{x+1}{x}}-1} \right| + C \\ &&&= \ln \left | \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}-\sqrt{x}} \right| + C \\ &&&= 2\ln \left | \sqrt{x+1}+\sqrt{x}\right| + C \\ &&&= 2\ln \left ( \sqrt{x+1}+\sqrt{x}\right) + C \\ \end{align*} \begin{align*} && V&= \pi \int_{1/8}^{9/16} y^2 \d x \\ &&&= \pi \int_{1/8}^{9/16} \left ( \frac1x + \frac{1}{x+1} - \frac{2}{\sqrt{x(x+1)}}\right) \d x \\ &&&= \pi \left [ \ln x + \ln (x+1) - 4 \ln(\sqrt{x+1} + \sqrt{x}) \right]_{1/8}^{9/16} \\ &&&= \pi \left ( \ln \frac{9}{16} + \ln \frac{25}{16} - 4 \ln \left ( \frac54 + \frac34\right) \right) +\\ &&&\quad -\pi \left ( \ln \frac{1}{8} + \ln \frac{9}{8} - 4 \ln \left ( \frac1{2\sqrt{2}} + \frac3{2\sqrt{2}}\right) \right) \\ &&&= \pi \left ( 2 \ln 3 - 8 \ln 2 + 2 \ln 5 - 4\ln2 \right) - \pi \left ( -6 \ln 2 + 2\ln 3 - 2\ln 2\right) \\ &&&= \pi (2 \ln 5 - 4 \ln 2 ) \\ &&&= 2 \pi \ln \tfrac54 \end{align*}

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Solution:

1. Notice that \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \text{Evaluate at }x = 1: && 2^n &= \sum_{i=0}^n \binom{n}{i} \end{align*}
2. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && n2^{n-1} &= \sum_{i=1}^n i\binom{n}{i} \end{align*}
3. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \Rightarrow && \int_0^1(1+x)^n \d x &= \int_0^1 \sum_{i=0}^n \binom{n}{i} x^i \d x \\ \Rightarrow && \frac{1}{n+1}(2^{n+1}-1) &= \sum_{i=0}^n \binom{n}{i}\int_0^1 x^i \d x\\ &&& = \sum_{i=0}^n \frac{1}{i+1}\binom{n}{i} \\ \end{align*}
4. \(\,\) \begin{align*} && (1+x)^n &= \sum_{i=0}^n \binom{n}{i} x^i \\ \frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i-1} \\ \times x: && nx(1+x)^{n-1} &= \sum_{i=1}^n i\binom{n}{i} x^{i} \\ \frac{\d}{\d x}: && n(1+x)^{n-1}+n(n-1)x(1+x)^{n-2} &= \sum_{i=1}^n i^2\binom{n}{i} x^{i-1} \\ \text{Evaluate at }x = 1: && \sum_{i=1}^n i^2\binom{n}{i} &= n(1+1)^{n-1}+n(n-1)x(1+1)^{n-2} \\ &&&= 2^{n-2} \left (n(n-1) + 2n \right) \\ &&&= n(n+1)2^{n-2} \end{align*}

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Solution: \begin{align*} && y &= e^x \\ && y' &= e^x \\ && y'' &= e^x \\ \Rightarrow && (x-1)y'' - x y' + y &= (x-1)e^x - xe^x + e^x \\ &&&= 0 \end{align*} Suppose \(y = ue^x\) then \begin{align*} && y' &= u'e^x + ue^x \\ && y'' &= (u''+u')e^x + (u'+u)e^x \\ &&&= (u''+2u' +u)e^x \\ \\ && 0 &= (x-1)y'' - x y' + y \\ &&&= [(x-1)(u''+2u'+u) - x(u'+u)+u]e^x \\ &&&= [(x-1)u'' +(x-2)u']e^x \\ \Rightarrow && 0 &= (x-1)u'' + (x-2)u' \\ v = u': && 0 &= (x-1)v' + (x-2) v \\ \Rightarrow && \frac{v'}{v} &= -\frac{x-2}{x-1} \\ &&&= -1-\frac{1}{x-1} \\ \Rightarrow && \ln v &= -x - \ln(x-1) + C \\ \Rightarrow && v &= A(x-1)e^{-x} \\ && u &= \int Axe^{-x} - Ae^{-x} \d x \\ &&&= \left [-Axe^{-x} +Ae^{-x} \right] + \int Ae^{-x} \d x \\ &&&= -Axe^{-x} + D\\ \Rightarrow && y &= ue^x \\ &&&= -Ax + De^x \end{align*}

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Solution:

1. \(\,\) \begin{align*} && I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\ u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\ &&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\ &&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\ \Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\ &&&= \int_0^a 1 \d x \\ &&&= a \\ \Rightarrow && I &= \frac{a}{2} \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\ &&&= \frac{1}{2} \tag{\(f(x) = \ln (x+1)\)} \\ \\ && K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\ &&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\ &&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\ &&&= \frac{\pi}{2\sqrt{2}} \end{align*}
2. \(\,\) \begin{align*} &&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\ u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\ &&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\ \Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\ &&&= \int_{\frac12}^2 \frac{1}{x} \d x\\ &&&= 2\ln2 \\ \Rightarrow && I &= \ln 2 \end{align*}

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Solution:

1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \int_{1/m}^m \frac{x^2}{x+1} \d x &= \int_{1/m}^m \left ( x- 1 + \frac{1}{x+1} \right) \d x \\ &&&= \left [ \frac{x^2}{2} - x + \ln (x+1) \right]_{1/m}^m \\ &&&= \left ( m^2/2 - m + \ln(m+1) \right)- \left ( \frac{1}{2m^2} - \frac{1}{m} + \ln\left(\frac1m+1\right) \right) \\ &&&= \frac{m^4-2m^3-1+2m}{2m^2} + \ln (m+1) - \ln(m+1) + \ln m \\ &&&= \frac{(m-1)^3(m+1)}{2m^2} + \ln m \end{align*}
2. \(\,\) \begin{align*} u = \frac{1}x, \d x = -\frac{1}{u^2} \d u:&& \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x &= \int_{u=m}^{u=1/m} \frac{1}{u^{-n}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/m}^m \frac{u^{n-1}}{u+1} \d u \end{align*}
3. • \(\bf (a)\) \(\,\) \begin{align*} && I &= \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \\ &&&= \int_{1/2}^2 \left ( \frac{x^2}{x+1} + \frac{3}{x^3(x+1)} \right) \d x \\ &&&= \int_{1/2}^2 \frac{x^2}{x+1} \d x + 3 \int_{1/2}^2 \frac{x^2}{x+1} \d x \\ &&&= 4 \left ( \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \right) \\ &&&= \frac32+4 \ln 2 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && J &= \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x \\ && K &= \int_1^2 \frac{x^5 +1}{x^3(x+1)}\, \d x\\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=1}^{u=1/2} \frac{u^{-5}+1}{u^{-3}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/2}^1 \frac{1 + u^5}{u^3(u+1)} \d u \\ \Rightarrow && K &= \frac12 \int_{1/2}^2 \frac{x^5+1}{x^3(x+1)} \d x \\ &&&= \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \\ &&&= \frac38 + \ln 2 \\ && L &= \int_1^2 \frac{x^3}{x^3(x+1)} \d x \\ &&&= \ln (3) - \ln 2 \\ \Rightarrow && J &= \frac38 + \ln 3 \end{align*}

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Solution:

1. \begin{align*} \mathcal{L}\{e^{-bt}f(t)\}(s) &= \int_0^{\infty}e^{-st}\{ e^{-bt}f(t) \} \d t \\ &= \int_0^{\infty} e^{-(s+b)t}f(t) \d t \\ &= F(s+b) \end{align*}
2. \begin{align*} \mathcal{L}\{f(at)\}(s) &= \int_0^{\infty} e^{-st}f(at) \d t \\ &= \int_{u=0}^{\infty}e^{-s \frac{u}{a}} f\left(a \tfrac{u}{a}\right)\frac{1}{a} \d u \\ &= \int_0^{\infty}e^{-su/a}f(u) a^{-1} \d u \\ &= a^{-1} \int_0^{\infty} e^{-(s/a)u}f(u) \d u \\ &= a^{-1} F\left (\frac{s}{a} \right) \end{align*}
3. \begin{align*} \mathcal{L}\{f'(t)\}(s) &= \int_0^{\infty} e^{-st}f'(t) \d t \\ &= \left [e^{-st} f(t) \right]_0^{\infty} - \int_0^{\infty} -s e^{-st} f(t) \d t\\ &= -f(0)+sF(s) \\ &= sF(s) - f(0) \end{align*}
4. Since \(f''(t) = -f(t)\) we must have: \begin{align*} && -\mathcal{L}(f)&= \mathcal{L}(f'') \\ &&&= s\mathcal{L}(f') -f'(0) \\ &&&= s(s\mathcal{L}(f)-f(0)) - f'(0) \\ &&&= s^2\mathcal{L}(f) - 1 \\ \Rightarrow && (1+s^2) \mathcal{L}(f) &= 1 \\ \Rightarrow && F(s) &= \frac{1}{1+s^2} \end{align*}

\begin{align*} \mathcal{L}\{e^{-pt}\cos qt\}(s) &= \mathcal{L}\{\cos qt\}(s+p) \\ &= q^{-1}\mathcal{L}\{\cos t\}\left (\frac{s+p}{q} \right) \\ &= q^{-1}\mathcal{L}\{\sin'\}\left (\frac{s+p}{q} \right) \\ &= q^{-1} \left (\frac{s+p}{q} \right) \mathcal{L}\{\sin\} \left (\frac{s+p}{q} \right) - q^{-1}\sin \left (0\right) \\ &= \frac{s+p}{q^2} \frac{1}{1+\left (\frac{s+p}{q} \right)^2 } \\ &= \frac{s+p}{q^2+(s+p)^2} \end{align*}

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Solution: \begin{align*} && t &= x+ \sqrt{x^2+2bx+c} \\ && \frac{\d t}{\d x} &= 1 + \frac{x+b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{x + \sqrt{x^2+2bx+c} + b}{\sqrt{x^2+2bx+c}} \\ &&&= \frac{t+b}{t-x} \\ \Rightarrow && \frac{\d x}{\d t} &= \frac{t-x}{t+b} \\ \\ && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{t-x} \frac{t-x}{t+b} \d t \\ &&&= \int \frac{1}{t+b} \d t \\ &&&= \ln (t + b) +C \\ &&&= \ln \left (x + \sqrt{x^2+2bx+c} + b \right) + C \end{align*} If \(b^2 = c\) then we have \(x^2+2bx+b^2 = (x+b)^2\) so \(\sqrt{x^2+2bx+c^2} = x+b\) (if \(x+b>0\)), so \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= \int \frac{1}{x+b} \d x\\ &&&= \ln (x + b) + C \\ &&&= \ln(x+b) + \ln 2 + C' \\ &&&= \ln (2(x+b)) + C' \\ &&&= \ln \left(x + b + \sqrt{(x+b)^2} \right)+C'\\ &&&= \ln \left(x + b + \sqrt{x^2+2bx+c} \right)+C'\\ \end{align*} If \(x+b < 0\) then the antiderivative is \(\ln 0\). \begin{align*} && \int \frac{1}{\sqrt{x^2+2bx+c}} \d x &= -\int \frac{1}{x+b} \d x\\ &&&= -\ln |x + b| + C \\ \end{align*} which are clearly different.

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Solution: \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{2 \sin x \cos x}{1+\sin^2x} \d x \\ &&&= \left [\ln (1 + \sin^2 x) \right]_0^{\pi/2} \\ &&&= \ln 2 \\ \\ && J &= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{2-\cos^2x} \d x \\ &&&= \frac{1}{2\sqrt{2}}\int_0^{\frac{1}{2}\pi} \left ( \frac{\sin x}{\sqrt{2}-\cos x}+ \frac{\sin x}{\sqrt{2}+\cos x} \right) \d x \\ &&&= \frac{1}{2\sqrt{2}} \left [\ln (\sqrt{2}-\cos x) - \ln (\sqrt{2}+\cos x) \right]_0^{\pi/2} \\ &&&= \frac{1}{2\sqrt{2}} \left (-\ln(\sqrt{2}-1)+\ln(\sqrt{2}+1) \right) \\ &&&= \frac1{2\sqrt{2}} \ln \left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)\\ &&&= \frac1{\sqrt{2}} \ln (\sqrt{2}+1) \end{align*} \begin{align*} && (1+\sqrt{2})^5 + (1-\sqrt{2})^5 &= 2(1+10\cdot2+5\cdot2^2) \\ &&&= 82 \\ && |(1-\sqrt{2})^5| & < 1 \\ && (1+\sqrt{2})^5 &< 83 < 99 \\ \\ && 1.4^2 &= 1.96 \\ &&&< 2 \\ \Rightarrow && 1.4 &<\sqrt{2} \\ \\ \Rightarrow && 2^{\sqrt{2}} &> 2^{1.4} \\ &&&=2^{7/5} \\ &&&= {128}^{1/5} \\ &&&>99^{1/5} \\ &&&>1+\sqrt{2} \end{align*} \begin{align*} && \ln 2 & > \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1) \\ \Leftrightarrow && \sqrt{2} \ln 2 &> \ln(\sqrt{2}+1) \\ \Leftrightarrow && 2^{\sqrt{2}} &> 1+\sqrt{2} \end{align*} which we have already shown, so the first integral is larger.

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Solution:

1. 

   + - ↺
   If \(y = \textrm{arcosh} x \), then \(\tanh\textrm{arcosh} x/2 = \sqrt{\frac{\cosh \textrm{arcosh} x-1}{\cosh \textrm{arcosh} x+1}} = \sqrt{\frac{x-1}{x+1}} = \frac{x-1}{\sqrt{x^2-1}}\)
2. \begin{align*} \int \textrm{arcosh} x \d x &= \left [x \textrm{arcosh} x \right] - \int \frac{x}{\sqrt{x^2-1}} \d x \\ &= x \textrm{arcosh} x - \sqrt{x^2-1}+C \\ \int \frac{x-1}{\sqrt{x^2-1}} &= \sqrt{x^2-1} - \textrm{arcosh} x +C \end{align*} Therefore \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} - 0 &> \sqrt{x^2-1} - \textrm{arcosh} x - 0 \\ \Rightarrow && (x+1) \textrm{arcosh} x &> 2\sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x & > 2\frac{\sqrt{x^2-1}}{x+1} \\ &&&= 2 \frac{\sqrt{x-1}}{\sqrt{x+1}} \\ &&&= 2 \frac{x-1}{\sqrt{x^2-1}} \end{align*}
3. Integrating both sides again, \begin{align*} && \int_1^x \textrm{arcosh} t \d t &> 2 \int_1^x \frac{t-1}{\sqrt{t^2-1}} \d t \\ \Rightarrow && x \textrm{arcosh} x - \sqrt{x^2-1} &> 2\left (\sqrt{x^2-1} - \textrm{arcosh}x \right) \\ \Rightarrow && (x+2)\textrm{arcosh} x &> 3 \sqrt{x^2-1} \\ \Rightarrow && \textrm{arcosh} x &> 3 \frac{\sqrt{x^2-1}}{x+2} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && x^2-y^2 &=(x-y)^3 \\ \Rightarrow && x+y &=d^2 \\ && x-y &= d \\ \Rightarrow && x &= \tfrac12(d^2+d) \\ && y &= \tfrac12(d^2-d) \end{align*} Therefore consider \(x^2 = m, y^2 = n\), so \(m = \tfrac14(d^2+d)^2, n = \tfrac14(d^2-d)^2\) so we want \(d^2-d > 20\), so \(d = 6, n = 225, m = 441\).
2. \(\,\) \begin{align*} && x^3-y^3 &= (x-y)^4 \\ \Rightarrow && x^2+xy+y^2 &= (x-y)^3 \\ && d^3 &= (x-y)^2+3xy \\ && d^3 &= d^2 + 3xy \\ \Rightarrow && 3xy &= d^3 - d^2 \\ \Rightarrow && 3x(x-d) &= d^3-d^2 \\ \Rightarrow && 0 &= 3x^2-3dx-(d^3-d^2) \\ \Rightarrow && 2x &=d \pm \sqrt{d^2+4\frac{(d^3-d^2)}{3}} \\ &&&= d \pm d \sqrt{\frac{3+4d-4}{3}} \\ &&&= d \pm d \sqrt{\frac{4d-1}{3}} \end{align*} Therefore we need \(\frac{4d-1}{3}\) to be an odd square. \(y = x-d = -\frac{d}{2} \pm \frac{d}{2} \sqrt{\frac{4d-1}{3}}\). Since we want positive values, we should take the positive square roots. \(d = \frac{3 \cdot 3^2 + 1}{4} = 7\) we have \(2x = 7 +7 \cdot 3 = 28 \Rightarrow x = 14, y = 7\)