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1

Quiz-5 / LEC0101 Quiz5

« on: October 31, 2019, 01:16:21 PM »

$\text{Find a particular solution of the given equation:}\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad ty^{\prime\prime}-(1+t)y^{\prime}+y=t^2e^{2t}, t>0;\qquad y_1(t)=1+t,  y_2(t)=e^t\\$
$\text{Write the given equation in standard form:}\\$
$y^{\prime\prime}-(\frac{1+t}{t})y^{\prime}+\frac{1}{t}y=te^{2t}\\$
$\text{Homogeneous equation is:}\\$
$y^{\prime\prime}-(\frac{1+t}{t})y^{\prime}+\frac{1}{t}y=0\\$
$\text{Verify $y_1(t)=1+t$ is the solution of the homogeneous equation}\\$
$y_1(t)=1+t, \quad y_1^{\prime}(t)=1,  \quad y_1^{\prime\prime}(t)=0\\$
$y_1^{\prime\prime}-(\frac{1+t}{t})y_1^{\prime}+\frac{1}{t}y_1=0-(\frac{1+t}{t})·1+\frac{1}{t}·(1+t)=0\\$
$\text{Therefore, $y_1(t)=1+t$ is a solution of the homogeneous equation}\\$
$\text{Verify $y_2(t)=e^t$ is the solution of the homogeneous equation}\\$
$y_2(t)=e^t, \quad y_2^{\prime}(t)=e^t, \quad y_2^{\prime\prime}(t)=e^t\\$
$y_2^{\prime\prime}-(\frac{1+t}{t})y_2^{\prime}+\frac{1}{t}y_2=e^t-(\frac{1+t}{t})·e^t+\frac{1}{t}·e^t=0\\$
$\text{Therefore, $y_2(t)=e^t$ is a solution of the homogeneous equation}\\$
$W=\begin{vmatrix}
1+t & e^t \\
1 & e^t \\
\end{vmatrix}=(1+t)(e^t)-1·(e^t)=e^t+te^t-e^t=te^t \neq 0\\$
$\text{Since $W\neq 0$ y_1 and y_2 form a fundamental set of solutions of the homogeneous equation}\\$
$\text{By using method of variation of parameters:}\\$
$Y(t)=-(1+t)\int{\frac{e^t·te^2t}{te^t}}dt+e^t\int{\frac{(1+t)·te^2t}{te^t}}dt\\$
$\qquad=-(1+t)\int e^{2t}dt+e^t\int(1+t)e^tdt\\$
$\qquad=-(1+t)\int e^{2t}dt+e^t[\int e^t dt+\int te^t dt]\\$
$\qquad=-(1+t)·\frac{1}{2}·e^{2t}+e^t[e^t+(te^t-\int e^tdt)]\\$
$\qquad=-(1+t)·\frac{1}{2}·e^{2t}+e^t[e^t+(te^t- e^t)]\\$
$\qquad=-(1+t)\frac{1}{2}e^{2t}+te^{2t}\\$
$\qquad=-\frac{1}{2}e^{2t}+\frac{1}{2}te^{2t}\\$
$\qquad=\frac{1}{2}(t-1)e^{2t}\\$
$\text{Hence, the particular solution is $Y(t)=\frac{1}{2}(t-1)e^{2t}$}$

2

Quiz-4 / TUT0401 Quiz 4

« on: October 18, 2019, 01:44:42 PM »

$9y^{\prime\prime}+6y^{\prime}+y=0\\$
$\text{We assume that $y=e^{rt}$ is a solution of this equation. Then the characteristic equation is }\\$
$9r^2+6r+1=0\\$
$\quad(3r+1)^2=0\\$
$\qquad\qquad r=-\frac{1}{3},-\frac{1}{3}\\$
$\text{Since the equation has two equal(repeated) roots.}\\$
$\text{Therefore, the general solution:}\\$
$y=c_1e^{-\frac{1}{3}t}+c_2te^{-\frac{1}{3}t}$

3

Quiz-3 / TUT0401 Quiz 3

« on: October 11, 2019, 01:59:57 PM »

$\text{Find the Wronskian of the given pair of functions}$
$\text{$cos^2(x)$, $1+cos(2x)$}$

$$\quad
   W= \begin{vmatrix}
     cos^2(x) & 1+cos(2x) \\
     -2cos(x)sin(x) & -2sin2x \\
    \end{vmatrix}
$$
$$   
= \begin{vmatrix}
     cos^2(x) & 1+cos(2x) \\
     -sin2x & -2sin2x \\
    \end{vmatrix}
$$
$$\qquad\qquad\qquad\quad
=-2cos^2(x)sin(2x)-(-sin(2x))(1+cos(2x))
$$
$$\qquad\quad
=(-sin2x)(2cos^2(x)-1-cos(2x))
$$
$$\qquad\qquad\quad
=(-sin2x)(2cos^2(x)-1-2cos^2(x)+1)
$$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
=(-sin2x)·0
\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
=0
$

4

Quiz-2 / TUT0401 Quiz2

« on: October 04, 2019, 02:00:12 PM »

$Question: (2xy^2+2y)+(2x^2y+2x)y^{\prime}=0\\$
$M=2xy^2+2y\qquad M_{y}=\frac{\partial}{\partial y}M=4xy+2\\$
$N=2x^2y+2x\qquad N_{x}=\frac{\partial}{\partial x}N=4xy+2\\$
$\text{Since $M_{y}=N_{x}$, the given differential equation is exact.}\\ $
$\text{$\exists \psi{(x,y)}$ such that $\psi_{x}=M=2xy^2+2y$}\\$
$\qquad\quad\psi{(x,y)}=\int {(2xy^2+2y)dx}\\$
$\qquad\qquad\qquad =x^2y^2+2xy+h(y)\\$
$\qquad\quad\psi_{y}=2x^2y+2x+h^{\prime}(y)\\$
$\text{set $\psi_{y}=N$}\\$
$\text{Therefore,}\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=0\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=x^2y^2+2xy=C\\$
$\text{Hence the solutions of the given differential equation are implicitly by}\\$
$\qquad\quad x^2y^2+2xy=C$

5

Quiz-1 / TUT0401 QUIZ1

« on: September 27, 2019, 02:15:53 PM »

$x{y^\prime } = {\left( {1 - {y^2}} \right)^{1/2}}.$


$\qquad$$\qquad$$\therefore Separable$

$\qquad$$\qquad$$\therefore x \frac{d y}{d x}=\sqrt{1-y^{2}}$

$\qquad$$\qquad$$Rearrange:~\int \frac{1}{\sqrt{1-y^{2}}} d y=\int \frac{1}{x} d x \quad~where~x \neq 0, y \neq \pm 1$

$\qquad\qquad {Integrating on both side}:$

$\qquad\qquad LHS:\int \frac{1}{\sqrt{1-y^{2}}}=\arcsin y=\ln |y|+C$

$\qquad\qquad RHS:\int \frac{1}{x} d x=\ln |x|$

$\qquad\qquad \therefore~General~sol:\arcsin y=\ln|x|+C$

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad y=\sin ( \ln |x |+C) \quad x \neq 0 \quad y \neq \pm 1$

6

Quiz-1 / TUT0401 Quiz 1

« on: September 27, 2019, 02:00:36 PM »

$\frac{{dy}}{{dx}} =  - \frac{{4x + 3y}}{{2x + y}}$

$\frac{{dy}}{{dx}} =  - \frac{{4 + \frac{{3y}}{x}}}{{1 + \frac{y}{x}}}$

$u = \frac{y}{x} \Rightarrow y = ux$

$\frac{{dy}}{{dx}} = \frac{{d(ux)}}{{dx}} = \frac{{du}}{{dx}}x + u \cdot 1$

$\frac{{du}}{{dx}}x + u =  - \frac{{4 + 3u}}{{2 + u}}$

$\frac{{du}}{{dx}}x =  - \frac{{4 + 3u}}{{2 + u}} - \frac{{2u + {u^2}}}{{2 + u}}$

$\frac{{du}}{{dx}}x = \frac{{ - \left( {{u^2} + 5u + 4} \right)}}{{u + 2}}$

$\frac{{du}}{{dx}}x = \frac{{ - (u + 1)(u + 4)}}{{u + 2}}$

$\int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = - \int {\frac{1}{x}} dx$

$LHS: \int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = \int {\frac{A}{{(u + 1)}}}  + \frac{B}{{(u + 4)}}du$

$ = \int {\frac{{A(u + 4) + B(u + 1)}}{{(u + 1)(u + 4)}}} du$

$ = \int {\frac{{Au + 4A + Bu + B}}{{(u + 1)(u + 4)}}} du$

$ = \int {\frac{{(A + B)u + (4A + B)}}{{(u + 1)(u + 4)}}} du$

$\left\{ {\begin{array}{*{20}{l}}
{A + B = 1}\\
{4A + B = 2}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{A = \frac{1}{3}}\\
{B = \frac{2}{3}}
\end{array}} \right.} \right.$

$\therefore \int {\frac{{u + 2}}{{(u + 1)(u + 4)}}} du = \int {\frac{1}{{3(u + 1)}}}  + \frac{2}{{3(u + 4)}}du =  - \int {\frac{1}{x}} dx$

$\frac{1}{3}\ln |u + 1| + \frac{2}{3}\ln |u + 4| =  - \ln |x| + C$

$\ln |u + 1| + \ln |u + 4 {|^2} =  - 3\ln |x| + 3C$

$\ln \left| {\frac{y}{x} + 1} \right| + \ln {\left| {\frac{y}{x} + 4} \right|^2} =  - 3\ln |x| + 3C$

$\ln \left| {\frac{{y + x}}{x}} \right| + \ln {\left| {\frac{{y + 4x}}{x}} \right|^2} =  - 3\ln |x| + 3C$

$\ln |y + x| - \ln |x| + \ln |y + 4x| - 2\ln |x| =  - 3\ln |x| + 3C$

$\ln |y + x| + \ln |y + 4x{|^2} = 3C$

${e^{\ln |y + x||y + 4x{|^2}}} = {e^{3C}}$

$|y + x||y + 4x{|^2} = C$