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### Messages - ZYR

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##### Term Test 2 / Re: Problem 1 (main sitting)
« on: November 19, 2019, 01:46:31 PM »
For the third solution, I believe there are some typo in Question a). For the last two equations, $y_2$ is missing, it should be $\frac{1}{2} \sin2t( 2t - tant)$, or simplify $\sin2t(t - \frac{1}{2}tant)$.

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##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 12:40:24 PM »
For question (b), I  think the first student who reply it forget to verify that $y_1(x) = xcos(x)$ is a solution.
Here is my solution: $y_1(x) = xcos(x)$
$y_1'(x) = cos(x) - xsin(x)$
$y_1''(x) = -sin(x) - sin(x) - xcos(x)$
Then substitute this to the original equation, we have
$(x^2y'' -2xy' + (x^2 + 2)y = 0$
$x^2( -sin(x) - sin(x) - xcos(x)) -2x(cos(x) - xsin(x)) +(x^2 +2） xcos(x) = -2x^2sin(x) - x^3cos(x) -2xcos(x) +2x^2sin(x) + x^3cos(x) + 2xcos(x) = 0$

So,  $y_1(x) = xcos(x)$ is a solution.

Also, there is a mistake when the first student do the integral part for the question b), it should be $\int sec^2 x dx = tanx + c$

We are looking for another solution, not for all other solutions. No mistake. V.I.

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##### Quiz-4 / Re: TUT0801
« on: October 19, 2019, 01:03:59 AM »
The question says that there is no need to change the variables, dose anyone can refer to one problem from section 3.1?

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##### Quiz-4 / TUT0801
« on: October 19, 2019, 12:49:32 AM »
$t^2y" + ty' + y = 0$

Let $x = lnt$, $t >0$, then $\frac{\partial x}{\partial t} = \frac{\partial }{\partial t}(lnt) = \frac{1}{t}$
Then we have $\frac{\partial y}{\partial t} = \frac{\partial y}{\partial x} \frac{1}{t}$
$\frac{\partial^2 y}{\partial t^2} = \frac{\partial }{\partial dt}(\frac{\partial y}{\partial x} \frac{1}{t}) = \frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x})$

When we substitute these to the original equation, we have :
$t^2 (\frac{1}{t^2}(\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x})) + t\frac{\partial y}{\partial x} \frac{1}{t} + y = 0$

$\frac{\partial^2 y}{\partial x} - \frac{\partial y}{\partial x} + \frac{\partial y}{\partial x} + y = 0$

$\frac{\partial^2 y}{\partial x} + y = 0$
Then we have a homogeneous equation, $y''+ y = 0$
$r^2 = -1$, $r = \pm i$, since $\lambda = 0$, $\mu = 1$
Then the general solution of this differential equation $y(x) = c_1 cos(x) + c_2 sin(x)$
And then substitute $x = lnt$, we get $y(t) = c_1 cos(lnt) + c_2 sin(lnt)$

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##### Quiz-3 / TUT0801
« on: October 11, 2019, 03:22:10 PM »
Verify that the function $y_1$ and $y_2$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?

$x^2y''-x(x+2)y' + (x+2)y = 0$, $x >0$, $y_1 = x$, $y_2 = xe^x$

$W(y_1, y_2)(x) = \det{\begin{vmatrix}y_1 & y_2\\ y_1' & y_2'\end{vmatrix}} = \det{\begin{vmatrix}x & xe^x \\ 1 & xe^x + e^x\end{vmatrix}} = x(xe^x + e^x) - xe^x = xe^x( x+1-1) = x^2e^x$

since $x > 0$ and $e^x \neq 0$,  so $W = x^2e^x \neq 0$.
Therefore, $y_1$ and $y_2$ constitute a fundamental set of solutions.

Verify :
1)Show $y_1 = x$ is one of the solutions
We know $y_1 = x$, $y_1' = 1$, $y_1'' = 0$, and then use these substitute in the equation, we have
LHS: $-x(x+2) + (x+2)x = 0$
RHS : 0
since LHS = RHS = 0, so $y_1 = x$ is one of the solutions

2) Show $y_2 = xe^x$ is one of the solutions
We know $y_2 = xe^x$, $y_2' = xe^x + e^x$, $y_2'' = e^x(x+2)$, also use thess substitute in the equation, we get
LHS: $x^2e^x(x+2)-x(x+2)(xe^x + e^x) + (x+2)xe^x = (x+2)e^x(x^2 -x(x+1) + x ) = (x+2)e^x(x^2 -x^2 - x + x ) = (x+2)e^x 0 = 0$
RHS : 0
since LHS = RHS = 0, so $y_2 = xe^x$ is one of the solutions.

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##### Quiz-2 / TUT0801
« on: October 06, 2019, 02:11:39 AM »

$1 + (\frac{x}{y} - \sin(y))y' = 0$

Rewrite the equation, we have: $1dx + (\frac{x}{y} - \sin(y))dy = 0$

Let $M = 1$, and $N = \frac{x}{y}- \sin(y)$

Then $M_y = 0$, and $N_x= \frac{1}{y}$, where $M_y \neq N_x$
So, it is not exact.

Let $R_1 = \frac{M_y - N_x}{M} = \frac{-1}{y}$

So $\mu = e^{-\int R_1 dy} = e^{\int \frac{1}{y} dy} = e^{\ln{y}} = y$

Then multiply $\mu = y$ on both sides, we have

$y dx + (x - y\sin(y)) dy = 0$

and now it is exact since $M_y = 1 = N_x$

So $\exists\ \psi(x,y)$, such that $\psi_x = M = y$

$\psi = \int M dx = \int y dx = yx + h(y)$

Then, we can get $\psi_y = x + h'(y)$

Also, we know $\psi_y = N = x - y\sin(y)$, which implies that

$h'(y) = - y\sin(y)$

So $h(y) = \int h'(y) dy = \int - y\sin(y) dy$

By integrating by parts, we can know

$h(y) = y\cos{y} -\sin{y} + c$

Therefore, we can have

$\psi = yx + y\cos{y} -\sin{y} = c$

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