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### Messages - Yuying Chen

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1
##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 05:43:09 AM »
$\text(a)\\$
$r^2+4=0\\$
$\qquad r=\pm2i\\$
$\therefore y(t)=c_1 \cos{2t}+c_2 \sin{2t}\\$
$W=\begin{vmatrix} \cos2t & \sin2t \\ -2\sin2t & 2\cos2t \\ \end{vmatrix}=2\\$
$W_1=\begin{vmatrix} 0 & \sin2t \\ 1 & 2\cos2t \\ \end{vmatrix}=-\sin2t\\$
$W_2=\begin{vmatrix} \cos2t & 0 \\ -2\sin2t & 1 \\ \end{vmatrix}=\cos2t\\$
$y_c(t)=\cos2t \int \frac{-\sin2s·\frac{1}{\cos^2{s}}}{2}ds+\sin2t \int \frac{\cos2s·\frac{1}{cos^2{s}}}{2}ds\\$
$\qquad = \frac{1}{2}\cos2t \int \frac{-2sins·coss}{cos^2{s}}ds+\frac{1}{2}\sin2t\int\frac{2cos^2{s}-1}{cos^2{s}}ds\\$
$\qquad =-\cos2t \int tans ds+ \frac{1}{2}\sin2t \int (2-\sec^2{s})ds\\$
$\qquad = \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$y(t)=c_1 \cos{2t}+c_2 \sin{2t}+ \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$\text(b)\\$
$y^{\prime}(t)=-2c_1\sin2t =2c_2\cos2t-2\sin2t·\ln\cos(t)-\cos2t·tant+\cos2t(2t- tant)+\frac{1}{2}\sin2t(2-\sec^2{t})\\$
$y(0)=c_1+\ln1=0 \implies c_1=0\\$
$y^{\prime}(0)=2c_2+(0-\tan0)=0 \implies c_2=0\\$
$\therefore y(t)=\cos2t·\ln (\cos t)+\frac{1}{2}\sin2t(2t-tant)$

2
##### Term Test 2 / Re: Problem 2 (main sitting)
« on: November 19, 2019, 05:42:54 AM »
$\text(a)\\$
$W=ce^{-\int p(t)dt}=ce^{-\int 4dt}=ce^{-4t}\\$
$\text(b)\\$
$r^3+4r^2+r-6=0\\$
$(r-1)(r^2+5r+6)=0\\$
$r=1, r=-2, r=-3\\$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}\\$
$W=\begin{vmatrix} e^t & e^{-2t} & e^{-3t}\\ e^t & -2e^{-2t} & -3e^{-3t}\\ e^t & 4e^{-2t} & 9e^{-3t}\\ \end{vmatrix}=-12e^{-4t}\\$
$\text{$\therefore c=-12$compare with (a)}\\$
$\text(c)\\$
$y_p(t)=Ate^t\\$
$y_p^{\prime}(t)=Ae^t+Ate^t\\$
$y_p^{\prime\prime}(t)=2Ae^t+Ate^t\\$
$y_p^{\prime\prime\prime}(t)=3Ae^t+Ate^t\\$
$3Ae^t+Ate^t+8Ae^t+4Ate^t+Ae^t+Ate^t-6Ate^t=12Ae^t\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad 12A=24\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad A=2$
$\therefore y(t)=c_1e^t+c_2e^{-2t}+c_3e^{-3t}+2te^t$

3
##### Term Test 2 / Re: Problem 4 (noon)
« on: November 19, 2019, 05:17:25 AM »
$\det(A-\lambda I)=0\\$
$\begin{vmatrix} 1-\lambda & 3 \\ -2 & -3-\lambda\\ \end{vmatrix}=(1-\lambda)(-3-\lambda)+6=0\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad \therefore \lambda=-1\pm \sqrt 2 i\\$
$\text{when$\lambda =-1+ \sqrt 2 i$},\\$
$\begin{pmatrix} 2-\sqrt 2 i & 3 \\ -2 & -2-\sqrt 2 i \end{pmatrix}= \begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}\\$
$e^{(-1+ \sqrt 2 i)t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}=e^{-t}\begin{pmatrix} 2+\sqrt 2 i \\ -2 \end{pmatrix}(\cos\sqrt2t+i\sin\sqrt 2t)\\$
$\qquad\qquad\qquad\qquad =e^{-t}\begin{pmatrix} 2\cos\sqrt 2t+2i\sin\sqrt2t+\sqrt2i\cos\sqrt2t-\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t-2i\sin\sqrt2t \end{pmatrix}\\$
$\qquad\qquad\qquad\qquad =e^{-t}i\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t\\ -2\cos\sqrt2t \end{pmatrix}\\$
$\therefore x(t)=c_1e^{-t}\begin{pmatrix} 2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\ -2\sin\sqrt2t \end{pmatrix}+c_2e^{-t}\begin{pmatrix} 2\cos\sqrt 2t -\sqrt2\sin\sqrt2t \\ -2\cos\sqrt2t \end{pmatrix}\\$

4
##### 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}$}$

5
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 02:32:30 PM »
$\text{(a)}\\$
$M=y+3y^2e^{2x}\qquad M_{y}=\frac{\partial}{\partial y}M=1+6ye^{2x}\\$
$N=1+2ye^{2x}\quad\quad N_{x}=\frac{\partial}{\partial x}N=4ye^{2x}\\$
$\text{Since$M_{y}\neq N_{x}$, the given differential equation is not exact.}\\$

$R_2=\frac{M_y-N_x}{N}=\frac{1+6ye^{2x}-4ye^{2x}}{1+2ye^{2x}}=\frac{1+2ye^{2x}}{1+2ye^{2x}}=1\\$
$\mu=e^{\int R_2dx}=e^{\int1dx}=e^x\\$
$(e^{x}y+3y^2e^{3x})+(e^x+2ye^{3x})y^{\prime}=0\\ \\$

$\text{$\exists \psi{(x,y)}$such that$\psi_{x}=M$}\\$
$\qquad\quad\psi{(x,y)}=\int {(e^{x}y+3y^2e^{3x})dx}\\$
$\qquad\qquad\qquad =e^xy+y^2e^{3x}+h(y)\\$
$\qquad\quad\psi_{y}=e^x=2ye^{3x}+h^{\prime}(y)=N\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=C\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=e^xy+y^2e^{3x}=C\\$

$\text{(b)}\\$
$\text{Since y(0)=1}\\$
$e^0·1+1^2·e^{3·0}=C\\$
$C=2\\$
$\text{Thus,}\\$
$e^xy+y^2e^{3x}=2\\$

6
##### Term Test 1 / Re: Problem 4 (main)
« on: October 23, 2019, 02:31:23 PM »
$\text{(a)}\\$
$r^2-6r+10=0\\$
$r=\frac{6\pm\sqrt{36-40}}{2}=\frac{6\pm2i}{2}=3\pm i\\$
$\text{Homogeneous Equation:$y_c(x)=c_1e^{3x}\cos x+c_2e^{3x}\sin x$}\\ \\$

$y^{\prime\prime}-6y^{\prime}+10y=2e^{3x}\\$
$Y_1(x)=Ae^{3x}\\$
$Y_1^{\prime}(x)=3Ae^{3x}\\$
$Y_1^{\prime\prime}(x)=9Ae^{3x}\\$
$9Ae^{3x}-18Ae^{3x}+10Ae^{3x}=2e^{3x}\\$
$\qquad(9A-18A+10A)e^{3x}=2e^{3x}\\$
$\qquad\qquad\qquad\qquad\quad A=2\\$
$Y_1(x)=2e^{3x}\\ \\$

$y^{\prime\prime}-6y^{\prime}+10y=39\cos x\\$
$Y_2(x)=B\cos x+C\sin x\\$
$Y_2^{\prime}(x)=-B\sin x+C\cos x\\$
$Y_2^{\prime\prime}(x)=-B\cos x-C\sin x\\$

$-B\cos x-C\sin x+6B\sin x-6C\cos x+10B\cos x+10C\sin x=39\cos x$
\begin{cases}
(-B-6C+10B)\cos x=39\cos x\\\
(-C+6B+10C)\sin x=0
\end{cases}
\begin{cases}
B=3\\
C=-2
\end{cases}
$Y_2(x)=3\cos x-2\sin x\\$
$\text{General Solution:}\\$
$y(t)=c_1e^{3x}\cos x+c_2e^{3x}\sin x+2e^{3x}+3\cos x-2\sin x\\$

$\text{(b)}\\$
$y^{\prime}=3c_1e^{3x}\cos x-c_1e^{3x}\sin x+3c_2e^{3x}\sin x+c_2e^{3x}\cos x+6e^{3x}-\sin x-2\cos x\\$
$y(0)=0\Rightarrow c_1+2+3=0\\$
$y^{\prime}(0)=0\Rightarrow 3c_1+c_2+6-2=0\\$
\begin{cases}
c_1=-5\\
c_2=11
\end{cases}

$\text{Thus,}\\$
$y(t)=-5e^{3x}\cos x+11e^{3x}\sin x+2e^{3x}+3\cos x-2\sin x$

7
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 02:28:35 PM »
$\text{(a)}\\$
$r^2-2r-3=0\\$
$(r+1)(r-3)=0\\$
$r=-1, r=3\\$
$\text{Homogeneous Equation:$y_c(x)=c_1e^{-x}+c_2e^{3x}$}\\ \\$

$\text{Since we know$\cosh{x}=\frac{e^{x}+e^{-x}}{2} = \frac{1}{2}e^x+\frac{1}{2}e^{-x}$}\\$
$\text{Therefore,}\\$
$y^{\prime\prime}-2y^{\prime}-3y=8e^x+8e^{-x}\\$
$Y(x)=Ae^x+Bxe^{-x}\\$
$Y^{\prime}(x)=Ae^x+Be^{-x}-Bxe^{-x}\\$
$Y^{\prime\prime}(x)=Ae^x-Be^{-x}-Be^{-x}+Bxe^{-x}\\$
$Ae^x-2Be^{-x}+Bxe^{-x}-2Ae^x-2Be^{-x}+2Bxe^{-x}-3Ae^x-3Bxe^{-x} = 8e^x+8e^{-x}\\$
\begin{cases}
(A-2A-3A)e^{x}=8e^{x}\\\
(-2B-2B)e^{-x}=8e^{-x}
\end{cases}
\begin{cases}
A=-2\\
B=-2
\end{cases}
$Y(x)=-2e^x-2xe^{-x}\\$
$\text{General Solution:}\\$
$y(t)=c_1e^{-x}+c_2e^{3x}-2e^x-2xe^{-x}\\$

$\text{(b)}\\$
$y^{\prime}=-c_1e^{-x}+3c_2e^{3x}-2e^{x}-2e^{-x}+2xe^{-x}\\$
$y(0)=0\Rightarrow c_1+c_2-2=0\\$
$y^{\prime}(0)=0\Rightarrow -c_1+3c_2-2-2=0\\$
\begin{cases}
c_1=\frac{1}{2}\\
c_2=\frac{3}{2}
\end{cases}

$\text{Thus,}$
$y(t)=\frac{1}{2}e^{-x}+\frac{3}{2}e^{3x}-2e^x-2xe^{-x}$

8
##### Term Test 1 / Re: Problem 1 (noon)
« on: October 23, 2019, 08:34:58 AM »
$\text{(a)}\\$
$M=2y+y^{2}\sin x\qquad M_{y}=\frac{\partial}{\partial y}M=2+2y\sin x\\$
$N=\sin 2x+2y\cos x\quad\quad N_{x}=\frac{\partial}{\partial x}N=2\cos 2x-2y\sin x\\$
$\text{Since$M_{y}\neq N_{x}$, the given differential equation is not exact.}\\$

$R_2=\frac{M_y-N_x}{N}=\frac{2+2y\sin x-2\cos 2x+2y\sin x}{\sin2 x+2y\cos x}=\frac{4\sin x(y+\sin x)}{2\cos x(y+\sin x)}=2\tan x\\$
$\mu=e^{\int R_2dx}=e^{2\int{\tan x}dx}=\sec^{2} x\\$
$(2y\sec^{2} x+y^{2}\sin x\ sec^{2} x)+(\sin 2x\sec^{2} x+2y\cos x\sec^{2} x)=0\\ \\$

$\text{$\exists \psi{(x,y)}$such that$\psi_{y}=N$}\\$
$\qquad\quad\psi{(x,y)}=\int{\sin 2x\sec^{2} x+2y\cos x\sec^{2} x}dy\\$
$\qquad\qquad\qquad =2\cos 2x\sec^{2} xy+y^2\sec^{2} x\cos x+h(x)\\$  OK. V.I.
$\qquad\quad\psi_{x}=2y\sec^{2} x\\$ ? V.I.
$\qquad\quad h^{\prime}(x)=y^{2}\sin x\sec^{2} x\\$
$\qquad\quad h(x)=y^{2}\sec x\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=2\cos2 x\sec^{2} xy+y^2\sec^{2} x\cos x+y^{2}\sec x=y\sec x(\sin 2x\sec^{2} x+2y)=C\\$

$\text{(b)}\\$   Wrong substitution. V.I.
$\text{Since$y(\frac{\pi}{4})=\sqrt{2}$}\\$
$C=6\sqrt{2}\\$
$\text{Thus,}\\$
$y\sec x(\sin 2x\sec^{2} x+2y)=6\sqrt{2}\\$

Wrong V.I.

9
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 07:10:28 AM »
$\text{(a)}\\$
$y^{\prime\prime}-\frac{2}{x}y^{\prime}+\frac{x^2+2}{x^2}y=0\\$
$W=ce^{-\int {p(x)dx}}=ce^{2\int {\frac{1}{x}dx}}=cx^2\\$
$\text{Let$c=1, W=x^2$}\\ \\$

$\text{(b)}\\$
$\text{Verify that$y_1(x) = x\cos x$is the solution}\\$
$y_1(x) = x\cos x\\$
$y_1^{\prime}(x) = \cos x-x\sin x\\$
$y_1^{\prime\prime}(x)= -\sin x-\sin x -x\cos x = -2\sin x-x\cos x\\$
$\text{Substitute these into the given equation:$x^2y^{\prime\prime}-2xy^{\prime}+(x^2+2)y=0$}\\$
$\quad x^{2}(-2\sin x-x\cos x)-2x(\cos x-x\sin x)+(x^{2}+2)(x\cos x)=0\\$
$-2x^{2}\sin x-x^{3}\cos x-2x\cos x+2x^{x}\sin x+x^{3}\cos x+2x\cos x=0\\$
$\text{Therefore,$y_1(x)=x\cos x$is the solution.}\\ \\$
$W=\begin{vmatrix} x\cos x & y_2 \\ \cos x-x\sin x & y_2^{\prime} \\ \end{vmatrix}=x^2\\$

$(x\cos x)y_2^{\prime}-(\cos x-x\sin x)y_2=x^2\\$
$y_2^{\prime}- \frac{\cos x-x\sin x}{x\cos x}y_2=\frac{x}{\cos x}\\$
$\mu = e^{\int {p(x)dx}}=e^{-\int {\frac{\cos x-x\sin x}{x\cos x}dx}} \qquad\text{By substitution,$u=x\cos x, du=(\cos x-x\sin x)dx$}\\$
$\qquad\qquad\quad =e^{-\int \frac{1}{u}}du\\$
$\qquad\qquad\quad=e^{-lnu}\\$
$\qquad\qquad\quad=\frac{1}{u}\\$
$\qquad\qquad\quad=\frac{1}{x\cos x}\\ \\$

$\frac{1}{x\cos x}y_2^{\prime}-{\frac{\cos x-x\sin x}{x^{2}\cos^{2} x} y_2}=\frac{1}{\cos^{2} x}\\$
$\frac{1}{x\cos x}y_2=\int \frac{1}{\cos^{2} x}dx\\$
$\qquad\quad=\int \sec^{2} xdx\\$
$\qquad\quad=\tan x+c\\$
$y_2=x\cos x·\tan{x}+x\cos x·c\\$
$\quad=x\cos x·\frac{\sin x}{\cos x}+x\cos x·c\\$
$\text{Let$c=0, y_2=x\sin x$}\\$
$\text{Therefore,}\\$
$y=c_1x\cos x+c_2x\sin x\\$

$\text{(c)}\\$
$y^{\prime}=c_1\cos x-c_1x\sin x+c_2\sin x+c_2x\cos x\\$
$y(\frac{\pi}{2})=1\Rightarrow c_1·(\frac{\pi}{2})·\cos (\frac{\pi}{2})+c_2·(\frac{\pi}{2})·\sin (\frac{\pi}{2})=1 \\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad c_2=\frac{2}{\pi}\\$
$y^{\prime}(\frac{\pi}{2})=0\Rightarrow -c_1·\frac{\pi}{2}·1+c_2·1=0\\$
$\quad\qquad\qquad\qquad\qquad\qquad c_1=\frac{4}{\pi^2}\\$
$\text{Thus,}\\$
$y=\frac{4}{\pi^2}x\cos x+\frac{2}{\pi}x\sin x$

No reason to post after this. V.I.

10
##### 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}$is a solution of this equation. Then the characteristic equation is }\\$
$9r^2+6r+1=0\\$
$\quad(3r+1)^2=0\\$
$\qquad\qquad r=-\frac{1}{3},-\frac{1}{3}\\$
$\text{Since the equation has two equal(repeated) roots.}\\$
$\text{Therefore, the general solution:}\\$
$y=c_1e^{-\frac{1}{3}t}+c_2te^{-\frac{1}{3}t}$

11
##### 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)$,$1+cos(2x)$}$

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

12
##### 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$}\\$
$\text{Therefore,}\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=0\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=x^2y^2+2xy=C\\$
$\text{Hence the solutions of the given differential equation are implicitly by}\\$
$\qquad\quad x^2y^2+2xy=C$

13
##### 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$

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

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

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

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

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

14
##### Quiz-1 / TUT0401 Quiz 1
« on: September 27, 2019, 02:00:36 PM »
$\frac{{dy}}{{dx}} = - \frac{{4x + 3y}}{{2x + y}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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