BE Civil Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2076 Nepal

Augmented matrix:

$$[A|B]=\begin{bmatrix} 1 & 2 & 3 & | & 14 \\ 2 & -1 & 5 & | & 15 \\ 3 & 1 & -2 & | & -1 \end{bmatrix}$$

Step 1. $R_2 \to R_2 - 2R_1$, $R_3 \to R_3 - 3R_1$:

$$\begin{bmatrix} 1 & 2 & 3 & | & 14 \\ 0 & -5 & -1 & | & -13 \\ 0 & -5 & -11 & | & -43 \end{bmatrix}$$

Here $R_2$: $(2-2,\,-1-4,\,5-6\,|\,15-28)=(0,-5,-1\,|\,-13)$ and $R_3$: $(3-3,\,1-6,\,-2-9\,|\,-1-42)=(0,-5,-11\,|\,-43)$.

Step 2. $R_3 \to R_3 - R_2$:

$$\begin{bmatrix} 1 & 2 & 3 & | & 14 \\ 0 & -5 & -1 & | & -13 \\ 0 & 0 & -10 & | & -30 \end{bmatrix}$$

since $(-11)-(-1)=-10$ and $(-43)-(-13)=-30$. This is row-echelon form.

Back substitution: From $R_3$: $-10z=-30 \Rightarrow z=3.$ From $R_2$: $-5y-z=-13 \Rightarrow -5y-3=-13 \Rightarrow -5y=-10 \Rightarrow y=2.$ From $R_1$: $x=14-2y-3z=14-4-9=1.$

Verification: Eq. 1: $1+2(2)+3(3)=1+4+9=14$ ✓ Eq. 2: $2(1)-2+5(3)=2-2+15=15$ ✓ Eq. 3: $3(1)+2-2(3)=3+2-6=-1$ ✓

Solution: $\boxed{x=1,\; y=2,\; z=3}$

Consistency condition (rank criterion): For a system $AX=B$ of $m$ equations in $n$ unknowns, the system is consistent if and only if $\operatorname{rank}(A)=\operatorname{rank}([A|B])=r$. If $r=n$ the solution is unique; if $r<n$ there are infinitely many solutions with $n-r$ free parameters. If $\operatorname{rank}(A)<\operatorname{rank}([A|B])$ the system is inconsistent (no solution).

(a) Symmetric equations. With point $(2,-1,3)$ and direction ratios $(1,2,-2)$:

$$\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-2}=t$$

Parametric form: $x=2+t,\;y=-1+2t,\;z=3-2t.$

(b) Intersection with plane $2x+y-z+5=0$. Substitute the parametric coordinates:

$$2(2+t)+(-1+2t)-(3-2t)+5=0$$ $$4+2t-1+2t-3+2t+5=0$$ $$6t+5=0 \Rightarrow t=-\frac{5}{6}$$

Then

$$x=2-\tfrac{5}{6}=\tfrac{7}{6},\quad y=-1+2(-\tfrac{5}{6})=-1-\tfrac{5}{3}=-\tfrac{8}{3},\quad z=3-2(-\tfrac{5}{6})=3+\tfrac{5}{3}=\tfrac{14}{3}.$$

Point of intersection: $\boxed{\left(\tfrac{7}{6},\,-\tfrac{8}{3},\,\tfrac{14}{3}\right)}.$

Check: $2(\tfrac{7}{6})+(-\tfrac{8}{3})-\tfrac{14}{3}+5=\tfrac{7}{3}-\tfrac{8}{3}-\tfrac{14}{3}+\tfrac{15}{3}=\tfrac{7-8-14+15}{3}=0$ ✓

(c) Distance from $Q(4,3,-1)$ to plane $2x+y-z+5=0$. Use

$$D=\frac{|2x_0+y_0-z_0+5|}{\sqrt{2^2+1^2+(-1)^2}}=\frac{|2(4)+3-(-1)+5|}{\sqrt{6}}=\frac{|8+3+1+5|}{\sqrt6}=\frac{17}{\sqrt6}.$$ $$D=\frac{17}{\sqrt6}=\frac{17\sqrt6}{6}\approx 6.94\text{ units}.$$

Perpendicular distance: $\boxed{D=\dfrac{17}{\sqrt6}\approx 6.94\text{ units}}.$

Complementary function (CF). Auxiliary equation:

$$m^2-5m+6=0 \Rightarrow (m-2)(m-3)=0 \Rightarrow m=2,\,3.$$ $$y_c = C_1 e^{2x} + C_2 e^{3x}.$$

Particular integral for $e^{2x}$. Since $m=2$ is a root of the auxiliary equation, the trial $Ae^{2x}$ fails; use $y_{p1}=A x e^{2x}$. Using the operator method, $P.I. = \dfrac{1}{D^2-5D+6}e^{2x}$. Because $D=2$ makes the denominator zero, multiply by $x$ and differentiate the denominator:

$$y_{p1}=x\cdot\frac{1}{2D-5}e^{2x}\Big|_{D=2}=x\cdot\frac{1}{2(2)-5}e^{2x}=x\cdot\frac{1}{-1}e^{2x}=-x e^{2x}.$$

So $y_{p1}=-x e^{2x}.$

Particular integral for $\sin x$. Replace $D^2$ by $-1$:

$$y_{p2}=\frac{1}{D^2-5D+6}\sin x=\frac{1}{-1-5D+6}\sin x=\frac{1}{5-5D}\sin x=\frac{1}{5}\cdot\frac{1}{1-D}\sin x.$$

Multiply numerator and denominator by $(1+D)$:

$$=\frac{1}{5}\cdot\frac{1+D}{1-D^2}\sin x=\frac{1}{5}\cdot\frac{(1+D)\sin x}{1-(-1)}=\frac{1}{5}\cdot\frac{\sin x+\cos x}{2}=\frac{1}{10}(\sin x+\cos x).$$

So $y_{p2}=\dfrac{1}{10}(\sin x+\cos x).$

General solution.

$$\boxed{y = C_1 e^{2x}+C_2 e^{3x}-x e^{2x}+\frac{1}{10}(\sin x+\cos x).}$$

Verification of $y_{p2}$: let $y=\tfrac1{10}(\sin x+\cos x)$. Then $y'=\tfrac1{10}(\cos x-\sin x)$, $y''=\tfrac1{10}(-\sin x-\cos x)$. Substitute: $y''-5y'+6y=\tfrac1{10}[(-\sin x-\cos x)-5(\cos x-\sin x)+6(\sin x+\cos x)]=\tfrac1{10}[(-1+5+6)\sin x+(-1-5+6)\cos x]=\tfrac1{10}[10\sin x+0]=\sin x$ ✓

Symmetry. $f(x)=x$ on $(-\pi,\pi)$ is an odd function, so $a_0=0$ and $a_n=0$ for all $n$. Only sine terms survive.

Fourier sine coefficients.

$$b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x\sin nx\,dx=\frac{2}{\pi}\int_0^{\pi}x\sin nx\,dx.$$

Integrate by parts with $u=x,\;dv=\sin nx\,dx$:

$$\int_0^{\pi}x\sin nx\,dx=\left[-\frac{x\cos nx}{n}\right]_0^{\pi}+\frac{1}{n}\int_0^{\pi}\cos nx\,dx.$$

The second integral $=\frac{1}{n}\left[\frac{\sin nx}{n}\right]_0^{\pi}=0$. The first term $=-\frac{\pi\cos n\pi}{n}=-\frac{\pi(-1)^n}{n}.$ Therefore

$$b_n=\frac{2}{\pi}\cdot\left(-\frac{\pi(-1)^n}{n}\right)=-\frac{2(-1)^n}{n}=\frac{2(-1)^{n+1}}{n}.$$

Fourier series.

$$\boxed{f(x)=x=2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx = 2\left(\sin x-\frac{\sin 2x}{2}+\frac{\sin 3x}{3}-\cdots\right).}$$

Deduction. Put $x=\dfrac{\pi}{2}$ (a point of continuity, $f(\pi/2)=\pi/2$):

$$\frac{\pi}{2}=2\left(\sin\frac{\pi}{2}-\frac{1}{2}\sin\pi+\frac{1}{3}\sin\frac{3\pi}{2}-\frac{1}{4}\sin 2\pi+\frac{1}{5}\sin\frac{5\pi}{2}-\cdots\right).$$

The sine values are $1,0,-1,0,1,0,-1,\dots$, so

$$\frac{\pi}{2}=2\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\right).$$

Dividing by 2:

$$\boxed{\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots}\quad\text{(Leibniz series).}$$

(a) Ratio test for $\sum \dfrac{n!}{n^n}$. Let $a_n=\dfrac{n!}{n^n}$.

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac1n\right)^n}.$$

As $n\to\infty$, $\left(1+\frac1n\right)^n\to e$, so

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}\approx 0.368<1.$$

Since the limit $<1$, by the ratio test the series converges. $\boxed{\text{Convergent}}$

(b) Power series $\sum \dfrac{(x-2)^n}{n\,3^n}$. Let $a_n=\dfrac{1}{n\,3^n}$. Radius of convergence:

$$\frac{1}{R}=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{n\,3^n}{(n+1)\,3^{n+1}}=\lim_{n\to\infty}\frac{n}{3(n+1)}=\frac{1}{3}.$$

Thus $R=3$. The series converges for $|x-2|<3$, i.e. $-1<x<5$.

Endpoint tests.

• At $x=5$: $|x-2|=3$, series $=\sum \dfrac{3^n}{n\,3^n}=\sum\dfrac{1}{n}$, the harmonic series, which diverges.
• At $x=-1$: $x-2=-3$, series $=\sum \dfrac{(-3)^n}{n\,3^n}=\sum\dfrac{(-1)^n}{n}$, the alternating harmonic series, which converges (Leibniz test).

Interval of convergence: $\boxed{[-1,\,5)}$ with radius $R=3$.

Determinant (expand along row 1).

$$\det A = 1\begin{vmatrix}1&4\\6&0\end{vmatrix}-2\begin{vmatrix}0&4\\5&0\end{vmatrix}+3\begin{vmatrix}0&1\\5&6\end{vmatrix}$$ $$=1(1\cdot0-4\cdot6)-2(0\cdot0-4\cdot5)+3(0\cdot6-1\cdot5)$$ $$=1(-24)-2(-20)+3(-5)=-24+40-15=1.$$ $$\boxed{\det A = 1}\;(\neq 0\Rightarrow A^{-1}\text{ exists}).$$

Cofactor matrix. Cofactor $C_{ij}=(-1)^{i+j}M_{ij}$:

$$C_{11}=+\begin{vmatrix}1&4\\6&0\end{vmatrix}=-24,\quad C_{12}=-\begin{vmatrix}0&4\\5&0\end{vmatrix}=-(-20)=20,\quad C_{13}=+\begin{vmatrix}0&1\\5&6\end{vmatrix}=-5,$$ $$C_{21}=-\begin{vmatrix}2&3\\6&0\end{vmatrix}=-(-18)=18,\quad C_{22}=+\begin{vmatrix}1&3\\5&0\end{vmatrix}=-15,\quad C_{23}=-\begin{vmatrix}1&2\\5&6\end{vmatrix}=-(-4)=4,$$ $$C_{31}=+\begin{vmatrix}2&3\\1&4\end{vmatrix}=5,\quad C_{32}=-\begin{vmatrix}1&3\\0&4\end{vmatrix}=-4,\quad C_{33}=+\begin{vmatrix}1&2\\0&1\end{vmatrix}=1.$$

Adjoint $=$ transpose of cofactor matrix:

$$\operatorname{adj}A=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}.$$

Inverse $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A=\dfrac{1}{1}\operatorname{adj}A$:

$$\boxed{A^{-1}=\begin{bmatrix}-24&18&5\\20&-15&-4\\-5&4&1\end{bmatrix}.}$$

Check (first entry of $AA^{-1}$): row1 of $A$ $\cdot$ col1 of $A^{-1}$ $=1(-24)+2(20)+3(-5)=-24+40-15=1$ ✓

Radius. The radius is the distance from centre $C(0,1,0)$ to the point $P(1,2,2)$ on the sphere:

$$r=\sqrt{(1-0)^2+(2-1)^2+(2-0)^2}=\sqrt{1+1+4}=\sqrt6.$$

Equation of sphere. Centre $(0,1,0)$, $r^2=6$:

$$\boxed{x^2+(y-1)^2+z^2=6}\quad\text{or}\quad x^2+y^2+z^2-2y-5=0.$$

Tangent plane at $P(1,2,2)$. The normal to the sphere at $P$ is the radius vector $\vec{CP}=(1-0,\,2-1,\,2-0)=(1,1,2)$. The tangent plane through $P$ with normal $(1,1,2)$ is

$$1(x-1)+1(y-2)+2(z-2)=0$$ $$x-1+y-2+2z-4=0$$ $$\boxed{x+y+2z-7=0.}$$

Check: $P(1,2,2)$ gives $1+2+4-7=0$ ✓, and the perpendicular distance from centre $(0,1,0)$ to this plane is $\dfrac{|0+1+0-7|}{\sqrt{1+1+4}}=\dfrac{6}{\sqrt6}=\sqrt6=r$ ✓ (plane is tangent).

Set up. $M=3x^2y+2xy+y^3$, $N=x^2+y^2$.

$$M_y=3x^2+2x+3y^2,\qquad N_x=2x.$$

Not exact since $M_y\ne N_x$.

Integrating factor (function of $x$).

$$\frac{M_y-N_x}{N}=\frac{(3x^2+2x+3y^2)-2x}{x^2+y^2}=\frac{3x^2+3y^2}{x^2+y^2}=3.$$

This depends on $x$ only (it is the constant $3$), so

$$\mu(x)=e^{\int 3\,dx}=e^{3x}.$$

Make exact. Multiply through by $e^{3x}$:

$$\underbrace{e^{3x}(3x^2y+2xy+y^3)}_{M^*}\,dx+\underbrace{e^{3x}(x^2+y^2)}_{N^*}\,dy=0.$$

Check: $N^*_x=e^{3x}(2x)+3e^{3x}(x^2+y^2)=e^{3x}(3x^2+2x+3y^2)$ and $M^*_y=e^{3x}(3x^2+2x+3y^2)$ — equal, so exact. ✓

Solve. Integrate $N^*$ with respect to $y$ (simpler):

$$F=\int e^{3x}(x^2+y^2)\,dy=e^{3x}\left(x^2 y+\frac{y^3}{3}\right)+g(x).$$

Then $F_x=e^{3x}\!\left(3x^2y+\tfrac{y^3}{}\cdot 1\right)$… compute carefully:

$$F_x=3e^{3x}\left(x^2y+\frac{y^3}{3}\right)+e^{3x}(2xy)+g'(x)=e^{3x}(3x^2y+y^3+2xy)+g'(x).$$

Set equal to $M^*=e^{3x}(3x^2y+2xy+y^3)$, giving $g'(x)=0\Rightarrow g(x)=$ const.

General solution.

$$\boxed{e^{3x}\left(x^2 y+\frac{y^3}{3}\right)=C.}$$

(a) $\displaystyle\sum \frac{2n+1}{n^3+n}$ — Limit comparison test. For large $n$, $\dfrac{2n+1}{n^3+n}\sim\dfrac{2n}{n^3}=\dfrac{2}{n^2}$. Compare with $b_n=\dfrac{1}{n^2}$ (a convergent $p$-series, $p=2>1$).

$$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(2n+1)/(n^3+n)}{1/n^2}=\lim_{n\to\infty}\frac{n^2(2n+1)}{n^3+n}=\lim_{n\to\infty}\frac{2n^3+n^2}{n^3+n}=2.$$

The limit is finite and non-zero ($=2$), and $\sum 1/n^2$ converges, so by the limit comparison test the series converges. $\boxed{\text{Convergent}}$

(b) $\displaystyle\sum \frac{1}{\sqrt{n^2+1}}$ — Limit comparison test. For large $n$, $\sqrt{n^2+1}\sim n$, so $\dfrac{1}{\sqrt{n^2+1}}\sim\dfrac{1}{n}$. Compare with $b_n=\dfrac1n$ (divergent harmonic series).

$$\lim_{n\to\infty}\frac{1/\sqrt{n^2+1}}{1/n}=\lim_{n\to\infty}\frac{n}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{\sqrt{1+1/n^2}}=1.$$

The limit is finite and non-zero ($=1$), and $\sum 1/n$ diverges, so by the limit comparison test the series diverges. $\boxed{\text{Divergent}}$

Half-range cosine series. On $0<x<L$ with $L=2$, we extend $f$ as an even function. The series is

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos\frac{n\pi x}{L},\qquad a_n=\frac{2}{L}\int_0^{L}f(x)\cos\frac{n\pi x}{L}\,dx.$$

Constant term. With $L=2$:

$$a_0=\frac{2}{2}\int_0^{2}x\,dx=\left[\frac{x^2}{2}\right]_0^2=2.$$

So $\dfrac{a_0}{2}=1.$

Cosine coefficients.

$$a_n=\frac{2}{2}\int_0^2 x\cos\frac{n\pi x}{2}\,dx=\int_0^2 x\cos\frac{n\pi x}{2}\,dx.$$

Integrate by parts ($u=x$, $dv=\cos\frac{n\pi x}{2}dx$, $v=\frac{2}{n\pi}\sin\frac{n\pi x}{2}$):

$$a_n=\left[\frac{2x}{n\pi}\sin\frac{n\pi x}{2}\right]_0^2-\frac{2}{n\pi}\int_0^2\sin\frac{n\pi x}{2}\,dx.$$

At $x=2$, $\sin n\pi=0$, so the boundary term vanishes. The remaining integral:

$$\int_0^2\sin\frac{n\pi x}{2}dx=\left[-\frac{2}{n\pi}\cos\frac{n\pi x}{2}\right]_0^2=-\frac{2}{n\pi}(\cos n\pi-1)=-\frac{2}{n\pi}((-1)^n-1).$$

Therefore

$$a_n=-\frac{2}{n\pi}\cdot\left(-\frac{2}{n\pi}((-1)^n-1)\right)=\frac{4}{n^2\pi^2}((-1)^n-1).$$

For even $n$, $(-1)^n-1=0\Rightarrow a_n=0$. For odd $n$, $(-1)^n-1=-2\Rightarrow a_n=-\dfrac{8}{n^2\pi^2}.$

Series.

$$\boxed{f(x)=x=1-\frac{8}{\pi^2}\sum_{n\,\text{odd}}\frac{1}{n^2}\cos\frac{n\pi x}{2}=1-\frac{8}{\pi^2}\left(\cos\frac{\pi x}{2}+\frac{1}{9}\cos\frac{3\pi x}{2}+\frac{1}{25}\cos\frac{5\pi x}{2}+\cdots\right).}$$

Cauchy-Euler substitution. Try $y=x^m$. Then $y'=mx^{m-1}$, $y''=m(m-1)x^{m-2}$. Substituting:

$$x^2\,m(m-1)x^{m-2}-3x\,m x^{m-1}+4x^m=0$$ $$\left[m(m-1)-3m+4\right]x^m=0.$$

Indicial (auxiliary) equation:

$$m(m-1)-3m+4=m^2-4m+4=(m-2)^2=0\Rightarrow m=2\ (\text{repeated}).$$

General solution (repeated root). For a double root $m$, the two independent solutions are $x^m$ and $x^m\ln x$:

$$y=(C_1+C_2\ln x)\,x^2.$$

Apply initial conditions. At $x=1$: $\ln 1=0$, so $y(1)=C_1\cdot 1=C_1=2\Rightarrow C_1=2.$ Differentiate:

$$y'=\frac{d}{dx}\left[(C_1+C_2\ln x)x^2\right]=C_2\cdot\frac{1}{x}\cdot x^2+(C_1+C_2\ln x)\cdot 2x=C_2 x+2x(C_1+C_2\ln x).$$

At $x=1$: $y'(1)=C_2\cdot 1+2\cdot 1\cdot(C_1+0)=C_2+2C_1=5.$ With $C_1=2$: $C_2+4=5\Rightarrow C_2=1.$

Particular solution.

$$\boxed{y=(2+\ln x)\,x^2.}$$

Check: $y(1)=(2+0)\cdot1=2$ ✓. $y'=x+2x(2+\ln x)$; $y'(1)=1+2(2)=5$ ✓.