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### Topics - Kun Zheng

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##### Quiz-5 / LEC5201 QUIZ5
« on: November 01, 2019, 02:03:36 PM »
Good afternoon,
The question I had on lecture was:
Find the general solution of the given differential equation.
$y''+9y=9sec^23t, 0<t<\pi/6$
For the homogeneous side y''+9y=0:
$r^2+9=0$
$r^2=-9$
$r_1=3i, r_2=-3i$
so the complementary solution is
$y_c(t)=Acos3t+Bsin3t$
For the non-homogeneous side:
$p(t)=0, q(t)=9, g(t)=9sec^23t$ are continuous on $0<t<\pi/6$
$W_{y1,y2}(t)= \begin{vmatrix} y_1(t) & y_2(t) \\ y'_1(t) & y'_2(t) \end{vmatrix}= \begin{vmatrix} cos3t & sin3t \\ -3sin3t & 3cos3t \end{vmatrix}=3$
Thus, $W_1(t)=-\int \frac{y_2(t)g(t)}{W_{1,2}(t)}dt=-\int \frac{sin3t9sec^23t}{3}dt=-3\int \frac{sin3t}{cos^23t}dt$
$=-3\int tan3tsec3tdt=-sec3t$
And, $W_2(t)=\int \frac{y_1(t)g(t)}{W_{1,2}(t)}dt=\int \frac{3cos3t/cos^23t}{3}dt$
$=\int 3sec3t dt=ln|sec3t+tan3t|$
(Be careful the integral of trigonometric here, honestly I made a mistake may lose some mark here on the lecture quiz, we must memorize them)
In conclusion, $Y(t)=y_c(t)+y_p(t)$
$Y(t)=Acos3t+Bsin3t+cos3t(-sec3t)+ln|sec3t+tan3t|sin3t$
$Y(t)=Acos3t+Bsin3t+ln|sec3t+tan3t|sin3t-1$
This is all, hope you have a good reading week!

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##### Quiz-4 / TUT0602 QUIZ4
« on: October 18, 2019, 11:17:59 PM »
Hi everyone for quiz 4 we need to get the original equation of
$y''+y'-6y=12e^{-2t}+12e^{3t}$

First, we get the homogeneous form:
$r^2+r-6=0$
$(r+3)(r-2)=0$
$r1=-3, r2=2$

Then we can get:
$y=C_1e^{-2t}+C_2e^{3t}$
$y'=-2C_1e^{-2t}+3C_2e^{3t}$
$y''=4C_1e^{-2t}+9C_2e^{3t}$

so this means:
$-6B-2B+4B=12$
$-6A+3A+9A=12$
Solve to get
$A=2, B=-3$

So we rewrite:
$2e^{3t}-3e^{-2t}$

Plug in to the solution:
$y=2e^{3t}-3e^{-2t}+C_1e^{2t}+C_2e^{-3t}$

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##### Quiz-3 / TUT0602 QUIZ3
« on: October 12, 2019, 02:01:37 AM »
Hi everyone, my question is to get the solution of y"-4y'=0, use the initial points of y'(-2)=1 and y(-2)=-1

First of all, we assume that y=e^(rt), and r must be a root of the characteristic equation.
Hence, we rewrite it as:
$r^2 -4r=0$
$r(r-4)=0$
$r_1=0, r_2=4$

Then we have the general structure as:
$y=C_1e^{r_1t}+C_2e^{r_1t}$
$y=C_1+C_2e^{4t}$

Derivative $y=C_1+C_2e^{4t}$
$y'=4C_2e^{4t}$

Use the initial values to plug in y'(-2)=1, y(-2)=-1
Got $C_2=e^8/4$
Then $y=C_1+e^{4t+8}/4$
Got $C_1=3/4$

Therefore, the initial equation is $y=3/4+e^{4t+8}/4$
Note: $y\rightarrow \infty, t\rightarrow \infty$

Correct me if I made a wrong solution or wrong question!
Have a good weekend!

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##### Quiz-2 / TUT0602 QUIZ2
« on: October 04, 2019, 02:31:08 PM »
Hi everyone, the problem is to prove (x^2)(y^3) + x(1+y^2)y'=0 is not exact, then multiply μ(x,y)=1/(xy^3) to solve the equation

First of all, we observe it follows M(x,y)+N(x,y)∂y/∂x = 0 form, we can get M(x,y)=(x^)(2y^3), N(x,y)= x(1+y^2)=x+xy^2
Then we get My=∂/∂y((x^2)(y^3))=3(x^2)(y^2) and Nx=∂/∂x(x(1+y^2))=1+y^2
As we see My is not equal to Nx, this is not an exact equation.

Second, we try multiplying the integrating factor μ(x,y)=1/(xy^3) on both sides: x+((1/y^3) + (1/y))∂y/dx=0
We do the first step again: M(x,y)=x and N(x,y)=1/y^3 + (1/y)
Derivative them: My=∂/∂y(x) =0 ; Nx=∂/∂x((1/y^3) + (1/y))=0
As we see My=Nx, now this is an exact equation.

Now we need to get the solution: ∫ M∂x+N∂x=(1/2)x^2 - (1/2)y^(-2)+ln|y|=2ln|y|+(x^2)-(1/y^2)=c

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##### Quiz-1 / TUT0602 QUIZ1
« on: September 27, 2019, 08:34:28 PM »
The quiz question I've done this morning is in the attached png, hope it helps.

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