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### Topics - Ruojing Chen

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##### Quiz-5 / Lec5101 quiz5
« on: November 01, 2019, 02:00:00 PM »
Find the general solution of the given differential equation.
$y''+4y=3csc(2t), o<t<\frac{\pi}{2}$

For homogeneous equation: y''+4y=0
$$r^2+4=0$$
$$r=\pm2i$$
$$y_c(t)=c_1cos2t+c_2sin2t$$

For non-homogeneous equation:y''+4y=3csc(2t)
$$p(t)=0,q(t)=4,g(t)=3csc(2t)$$
$$W=2cos^2t+2sin^2t=2$$
$$u_1=-\int\frac{y_2(t)g(t)}{W}=-\int\frac{sin2t*3csc2t}{2}dt=-\frac{3}{2}t$$
$$u_2=\int\frac{y_1(t)g(t)}{W}=\int\frac{cos2t*3csc2t}{2}dt=\frac{3}{4}ln|2t|$$
$$y_p(t)=u_1y_1(t)+u_2y_2(t) =cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$

$$\therefore y=c_1cos2t+c_2sin2t+cos2t*(-\frac{3}{2}t)+sin2t*(\frac{3}{4}ln|2t|)$$

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##### Quiz-4 / tut 0202 quiz4
« on: October 18, 2019, 02:06:36 PM »
Find the general solution for equation y''-2y'+6y=0

$$set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$

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##### Quiz-4 / tut 0202 quiz4
« on: October 18, 2019, 02:02:34 PM »
Find the general solution for equation y''-2y'+6y=0

$$set \ r^2-2r+6=0$$

$$r=\frac{2\pm\sqrt{(-2)^2-4*6}}{2}$$

$$r=\frac{2\pm2i\sqrt{5}}{2}$$

$$r=1\pm\sqrt{5}i$$

$$y_c(t)=c_1e^tCos(\sqrt{5}t)+c_2e^tSin(\sqrt{5}t)$$

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##### Quiz-3 / TUT 0202 Quiz3
« on: October 11, 2019, 02:00:02 PM »
0202 solution

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##### Quiz-2 / TUT 0202 Quiz2
« on: October 04, 2019, 03:35:10 PM »

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##### Quiz-1 / TUT 0202 Quiz
« on: September 27, 2019, 03:00:21 PM »
solution

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##### Quiz-1 / TUT 0202 Quiz
« on: September 27, 2019, 02:11:52 PM »
dy/dx = ((x^2)+(3y^2))/(2xy)

dy/dt = (1+3y^2/x^2)/(2y/x)

let u = y/x, then y = ux

dy/dx = d(ux)/dx = (du/dx)*x + u = (1+3u^2) / (2u) = 1/(2u) + 3u/2
(du/dx)*x= 1/(2u) + u/2 = (1+u^2)/(2u)
∫ ( (2u) / (1+u^2) ) du = ∫ (1/x) dx
ln|1+(y/x)^2| = ln|x| + c

1+u = ax, a = e^c
1 + (y^2)/(x^2) - ax = 0
y^2 + x^2 - ax^3 = 0

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