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### Messages - yueyangyu

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1
##### Term Test 2 / Re: Problem 4 (main sitting)
« on: November 19, 2019, 12:52:21 PM »
4) $x'=\left(\begin{array}{cc} 3 & 3\\ -2 & -1 \end{array}\right)x$

a)

$det(A-\lambda I)=det(\begin{array}{cc} 3-\lambda & 3\\ -2 & -1-\lambda \end{array})=(3-\lambda)(-1-\lambda)+6=0$

$\lambda^{2}-2\lambda+3=0$

$(\lambda-1)^{2}=-2$

$\lambda=1\pm\sqrt{2}i$

$\lambda=1+\sqrt{2}i$

$(\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ -2 & -2-\sqrt{2}i & 0 \end{array}) = (\begin{array}{ccc} 2-\sqrt{2}i & 3 & 0\\ 0 & 0 & 0 \end{array})$

$v=(\begin{array}{c}2+\sqrt{2}i\\-2\end{array})$

$e^{(1+\sqrt{2}i)t}(\begin{array}{c} 2+\sqrt{2}i\\ -2 \end{array})=e^{t}(cos\sqrt{2}t+isin\sqrt{2}t)(\begin{array}{c}2+\sqrt{2}i\\-2\end{array}) = e^{t}(\begin{array}{c} 2cos\sqrt{2}t+2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t-2isin\sqrt{2}t \end{array})=e^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+ie^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$

$x(t)=c_{1}e{}^{t}(\begin{array}{c} 2cos\sqrt{2}t-\sqrt{2}sin\sqrt{2}t\\ -2cos\sqrt{2}t \end{array})+c_{2}e{}^{t}(\begin{array}{c} 2isin\sqrt{2}t+\sqrt{2}icos\sqrt{2}t\\ -2isin\sqrt{2}t \end{array})$

2
##### Quiz-5 / quiz5
« on: November 01, 2019, 02:40:58 AM »
Verify  that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
$$(1-t)y''+ty'-y=2(t-1)^2e^{-t}, 0<t<1;y1(t)=e^t \quad y2(t)=t$$

$$y1(t)=e^t \quad y'1(t)=e^t \quad y''1(t)=e^t$$
$$y2(t)=t \quad y'2(t)=1 \quad y''2(t)=0$$
Substitude back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
Verified that y1(t) and y2(t) both satisfy the corresponding homogeneous equation.
$$y_c(t)=C_1e^t+C_2t$$
Divide both side by 1-t, then
$$p(t)=\frac{t}{1-t} \quad q(t)=-\frac{1}{1-t} \quad g(t)=-2(t-1)e^{-t}$$
$$W(t)=(1-t)e^t$$
$$u1(t)=-\int{\frac{y2(t)g(t)}{W(t)}}dt=-2\int{te^{2t}}=(t+\frac{1}{2})e^{-2t}$$
$$u2(t)=\int\frac{y1(t)g(t)}{W(t)}dt=2\int{e^{-t}}=-2e^{-t}$$
Therefore, the particular solution is:
$$Y(t)=u1(t)y1(t)+u2(t)y2(t)=(t+\frac{1}{2})e^{-2t}*e^t+(-2e^{-t})*t=(\frac{1}{2}-t)e^{-t}$$
Hence, the general solution:
$$y(t)=y_c(t)+Y(t)=C_1e^t+C_2t+(\frac{1}{2}-t)e^{-t}$$
The particular solution is:
$$Y(t)=(\frac{1}{2}-t)e^{-t}$$

3
##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 04:56:50 PM »
$$M_{y}=1+6ye^{3x}$$
$$N_{x}=4ye^{2x}$$
$M_{y} ≠N_{x}$,it is not exact
$$R_{2} =\frac{ M_{y} -N_{x}}{N}=\frac{1+2ye^{2x} }{1+2ye^{2x}}=1$$
$$μ=e^{∫R_{2}dx} =e^{∫1 dx} =e^x$$
Multiplying both sides by $\mu$, we get
$$ye^x+3y^2e^{3x} +(e^x+2ye^{3x}) y^\prime=0$$
$$M_{y}^\prime=e^x+6ye^{3x}$$
$$N_{x}^\prime=e^x+6ye^{3x}$$
$M_{y}^\prime=N_{x}^\prime$,it is exact
$$∃φ(x,y) such that\ φ_{x} =M^\prime,φ_{y} =N^\prime$$
$$φ(x,y)=∫{M^\prime dx}=∫{ye^x+3y^{2}e^{3x}dx}=ye^x+y^{2}e^{3x} +h(y)$$
$$φ_{y} =e^x+2ye^{3x} +h(y)^\prime=e^x+2ye^{3x}$$
$h(y)^\prime=0$

So h(y)=c
$$φ(x,y)=ye^x+y^{2}e^{3x} =c$$
Since y(0)=1
$$1⋅e^0+1^2⋅e^0=2=c$$
$$φ(x,y)=ye^x+y^{2}e^{3x} =2$$

4
##### Term Test 1 / Re: Problem 2 (main)
« on: October 23, 2019, 04:49:43 PM »
a)
$$y'' - \frac{2}{x}y' + (1 + \frac{2}{x^2}) = 0$$

$$W(y_1, y_2)(x) = ce^{-\int p(x)dx} = ce^{\int(\frac{2}{x})dx} = ce^{2lnx} = cx^2$$

b)

Since $y_1(x) = xcosx$, $y'_1(x) = cosx-xsinx$, $y''_1(x) = -sinx-xcosx-sinx = -xcosx-2sinx$

$x^2y'' -2xy' + (x^2+2)y = 0 = -x^3cosx-2x^2sinx - 2xcosx + 2x^2sinx + x^3cosx + 2xcosx = 0$

So $y_1(x) = xcosx$ is a solution.

Let $c = 1, W(y_1, y_2)(x) = x^2$.

$$y_2 = y_1\int(\frac{ce^{-\int p(x)dx}}{(y_1)^2})dx = xcosx\int(\frac{x^2}{(x^2cos^2x})dx = xcosx\int sec^2xdx = xcosx tanx = xsinx$$

c)
The general solution is $y = c_1xcosx +c_2xsinx$

$y' = c_1(cosx -xsinx) + c_2(xcosx + sinx)$

$y(\frac{\pi}{2}) = 1, y'(\frac{\pi}{2}) = 0$, we have

$1 = \frac{\pi}{2}c_2$ and $0 = (-\frac{\pi}{2})c_1 + c_2$

So $c_2 = \frac{2}{\pi}$ and $c_1 = \frac{4}{(\pi)^2}$

$$y = \frac{4}{(\pi)^2}xcosx +\frac{2}{\pi}xsinx$$

5
##### Term Test 1 / Re: Problem 3 (main)
« on: October 23, 2019, 04:45:23 PM »
a)
First solve the homogenous part:
\begin{aligned}
r^{2}-2r-3 &=0\\
r^{2}-2r+1&=4\\
(r-1)^{2} &=4\\
r_1 &=3 \\r_2 &=-1
\end{aligned}
So the solution to homogenous part is:
\begin{aligned}
y_c(x) =c_1e^{3x}+c_2e^{-x}
\end{aligned}
Next we solve
\begin{aligned}
y^{\prime\prime}-2y^{\prime}-3y &=16cosh(x)\\
&=16*\frac{e^{x}+e^{-x}}{2} \\
&=8e^{x}+8e^{-x}
\end{aligned}
Let
\begin{aligned}
y_p(x)&=Ae^{x}+Be^{-x}*x\\
y_p^{\prime}(x) &= Ae^{x}+Be^{-x}-Bxe^{-x}\\
y_p^{\prime\prime}(x) &= Ae^{x}-Be^{-x}-Be^{-x}+Bxe^{-x}\\
&= Ae^{x}-2Be^{-x}+Bxe^{-x}\\
\end{aligned}
Thus we can have
\begin{aligned}
Ae^{x}-2Be^{-x}+Bxe^{-x}-2(Ae^{x}+Be^{-x}-Bxe^{-x})-3(Ae^{x}+Be^{-x}*x) &=8e^{x}+8e^{-x}\\
-4Ae^{x}-4Be^{x} &=8e^{x}+8e^{-x}\\
A &=-2\\
B &=-2
\end{aligned}
Therefore, we can have:
\begin{aligned}
y_p(x) &=-2e^{x}-2xe^{-x}
\end{aligned}
From the above, we get
\begin{aligned}
y&=y_c(x)+y_p(x)\\
&= c_1e^{3x}+c_2e^{-x}-2e^{x}-2xe^{-x}
\end{aligned}

b)
\begin{aligned}
y &= c_1e^{3x}+c_2e^{-x}-2e^{x}-2xe^{-x}\\
y_p^{\prime} &= 3c_1e^{3x}-c_2e^{-x}-2e^x+2xe^{-x}-2e^{-x}\\
\end{aligned}
since
\begin{aligned}
y(0) &=0\\
y^{\prime} &= 0
\end{aligned}
we can have
\begin{aligned}
c_1 &=3/2\\
c_2 &=1/2
\end{aligned}
Thus, the solution is
\begin{aligned}
y &= 3/2e^{3x}+1/2e^{-x}-2e^{x}-2xe^{-x}
\end{aligned}

6
##### Quiz-4 / quiz4 tut0502
« on: October 18, 2019, 04:05:06 PM »
Find the general solution of the differential equation
$y''+2y'+2y=0$

The characteristic equation of the given equation is:

$r^2+2r+2=0$

$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm{i}$

Then,
$$r_1=-1+i \quad r_2=-1-i$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{-t}cost+c_2e^{-t}sint$$

7
##### Quiz-3 / quiz3 tut0502
« on: October 11, 2019, 02:00:02 PM »
Find the general solution of the given differential equation.
$$y''-2y'-2y=0$$

Assume that $$y=e^{rt}$$ and it follows that r must be a root of characteristic equation
$$r^2-2r-2=0$$
$$r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Then,
$$r_1=1+\sqrt{3} \quad r_2=1-\sqrt{3}$$

Therefore, the general solution of the given differential equation is:
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

8
##### Quiz-2 / quiz2 0502
« on: October 04, 2019, 02:52:24 PM »
Find an integrating factor and solve the given equation:
$$(3x+\frac{6}{y})+(\frac{x^{2}}{y}+3\frac{y}{x})\frac{dy}{dx}=0$$

Firstly, find an integrating factor $\mu$ as a function of xy s.t.
$$(\mu M)_y=(\mu N)_x$$
Let $$z=xy$$.
$$\mu M_y+xM\frac{d\mu}{dz}=\mu N_x+yN\frac{d\mu}{dz}$$

$$\frac{d\mu}{dz}=\mu(\frac{N_x-M_y}{xM-yN})$$
Therefore,
$$\mu(z)=exp(\int(R(z))dz)$$
where $$R(z)=R(xy)=\frac{N_x-M_y}{xM-yN}$$
Let $$M=3x+\frac{6}{y}$$ $$N=\frac{x^{2}}{y}+3\frac{y}{x}$$
Then,
$$M_y=\frac{-6}{y^{2}} \quad N_x=\frac{2x}{y}-\frac{37}{x^{2}}$$
We can see that this equation is not exact
$$\frac{N_x-M_y}{xM-yN}=\frac{\frac{2x}{y}-\frac{3y}{x^{2}}+\frac{6}{y^{2}}}{2x^{2}+\frac{6x}{y}-\frac{3y^{2}}{x}}=\frac{1}{xy}$$
Thus, we have an integrating factor
$$\mu(xy)=exp(\int\frac{1}{z}dz)=z=xy$$
Multiplying the original equation by the integrating factor, we have
$$(3x^{2}y+6x)+(x^{3}+3y{2})\frac{dy}{dx}=0$$
This equation is exact because
$$M_y=N_x=3x^{2}$$
Thus, there exists a function $$\Psi(x,y)$$ such that $$\Psi_x(x,y)=3x^{2}y+6x$$
$$\Psi_y(x,y)=x^{3}+3y{2}$$
$$\Psi(x,y)=x^{3}y+3x^{2}+h(y)$$
Differentiating with respect to y, we get
$$\Psi_y(x,y)=x^{3}+h'(y)$$
$$h'(y)=3y^{2} \quad h(y)=y^{3}$$
and we have
$$\Psi(x,y)=x^{3}y+3x^{2}+y^{3}$$
Thus the solutions of the differential equation are given implicitly by
$$x^{3}y+3x^{2}+y^{3}=C$$

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