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### Messages - Darren Zhang

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16
##### Quiz-3 / Re: Q3-T0501
« on: February 10, 2018, 05:57:47 PM »
Substitution of the assumed solution $y=e^{rt}$ results in the characteristic equation $$2r^2-3r+1=0$$
The roots of the equation are $r = \frac{1}{2}, 1$. Hence the general solution is $y = c_{1}e^{\frac{t}{2}}+c_{2}e^{\frac{3t}{2}}$

17
##### Quiz-3 / Re: Q3-T0701
« on: February 10, 2018, 05:54:40 PM »
The characteristic equation is $r^2-2r-2=0$, with root of $r = 1+\sqrt{3}, 1-\sqrt{3}$.
Hence, the general solution should be $y = c_{1}exp(1-\sqrt{3})t+c_{2}(1+\sqrt{3})t$

18
##### Quiz-3 / Re: Q3-T0201
« on: February 10, 2018, 05:53:23 PM »
Let y = e^{rt},
Substitution of the assumed solution results in the characteristic equation $$r^2+3r+2=0$$
The roots of the equation are r = -2, -1 . Hence the general solution is $y = c_{1}e^{-t}+c_{2}e^{-2t}$

19
##### Quiz-3 / Re: Q3-T0101
« on: February 10, 2018, 05:48:41 PM »
Divide both side by $t^2$, $$y''-\frac{t+2}{t}y'+\frac{t+2}{t^2}=0$$
The Wronskian of two solutions can be calculated by the formula,
$$c(exp[-\int \frac{t+2}{t}dt])$$
Then we can get the solution
$$c*(e^t*t^2)$$

20
##### Quiz-1 / TUT 0601
« on: January 26, 2018, 01:18:34 PM »
Question
Find the general solution of the given function.
$$\frac{dy}{dx} = -\frac{(4x+3y)}{(2x+y)}$$

Let y = xv,
$\frac{dy}{dx} = v + v'x$
$v+v'x = -\frac{4x+3xv}{2x+xv}$
$v+v'x = -\frac{(4+3v)}{(2+v)}$
$v'x = \frac{-v^2-5v-4}{2+v}$
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{x} dx$
Let u = $(v^2+5v+4)$, du = (2v+5)dv
Then,
$\int \frac{2+v}{(v+1)(v+4)}dv = \int \frac{1}{2} \frac{(2v+5)}{(v^2+5v+4)}dv - \int \frac{0.5}{(v^2+5v+4)}dv$
$\frac{1}{2} ln(v^2+5v+4)dv - \frac{1}{2}*\frac{1}{3} \int(\frac{1}{v+1}-\frac{1}{v+4})dv$
$3ln(v^2+5v+4)-ln(\frac{(v+1)}{(v+4)}) = c-6ln(x)$
$(v+1)^2(v+1)^4 = x^6$
Therefore, we can get to know that
$(v+4)^2(v+1) = \frac{c}{x^{3}}$
plug y = xv into this equation
Therefore, we can get
$$(4x+y)^{2} (x+y) = c$$

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