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### Messages - Yue Sagawa

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1
##### Term Test 2 / Re: Problem 4 (morning)
« on: November 19, 2019, 06:18:55 AM »
I found the general real solution.

2
##### Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 07:34:01 AM »
Should the homogeneous solution be $𝑦_𝑐(𝑥)=𝑐_1𝑒^{3𝑥}\cos 4𝑥 + 𝑐_2𝑒^{3𝑥}\sin 4𝑥$?

3
##### Term Test 1 / Re: Problem 3 (noon)
« on: October 23, 2019, 06:49:12 AM »
Solution:
$$96 \sinh x = 96 \frac{e^{x}-e^{-x}}{2} = 48e^{x}-48e^{-x}$$
$$y" -4y'+3y = 48e^{x}-48e^{-x}$$
(a). Homogeneous part:
$$y"-4y'+3y=0$$
$$r^2-4r+3=0$$
$$(r-1)(r-3)=0$$
$$r_1 = 1, r_2=3$$
$$y_0 = c_1e^{x}+c_2e^{3x}$$

Next we solve $y" -4y'+3y = 48e^{x}$
Since we already have $e^{x}$ in our solution, let $y_1 = Axe^{x}$
Then $$y_1'=Axe^{x}+Ae^{x}$$
$$y_1" = 2Ae^{x}+Axe^{x}$$
$$(2Ae^{x}+Axe^{x})-4(Axe^x+Ae^{x})+3Axe^{x} = 48e^{x}$$
$$(2A-4A)e^{x}=48e^{x}$$
$$-2A = 48$$
$$A=-24$$
$$y_1=-24xe^{x}$$

Next we solve $y" -4y'+3y = -48e^{-x}$
Let $y_2=Be^{-x}$
Then
$$y_2' = -Be^{-x}$$
$$y_2" = Be^{-x}$$
$$Be^{-x} +4Be^{-x}+3Be^{-x}=-48e^{-x}$$
$$8B = -48$$
$$B = -6$$
$$y_2= -6e^{-x}$$
So the general solution is $y = y_0+y_1+y_2 = c_1e^{x}+c_2e^{3x} -24xe^{x}-6e^{-x}$

(b). $$y'=c_1e^{x}+3c_2e^{3x} -24e^{x}-24xe^x+6e^{-x}$$
$$y(0) = c_1+c_2-6 = 0 \Rightarrow c_1+c_2 = 6$$
$$y'(0) = c_1+3c_2-24+6 = 0 \Rightarrow c_1+3c_2=18$$
We get $$c_1 = 0, c_2=6$$
So $$y = 6e^{3x} -24xe^{x}-6e^{-x}$$

4
##### Quiz-4 / TUT0702 Quiz4
« on: October 18, 2019, 02:37:11 PM »
Question:
Find the solution of the given initial value problem.
$$y"+4y'+5y = 0, \ y(0)=1, \ y'(0) = 0$$
Solution:
The characteristic equation of the given differential equation is $r^2 + 4r+5 = 0$.
$$r = \frac{-4 \pm \sqrt{16-20}}{2}$$
$$r = -2 \pm i$$
Therefore, the general solution is $y = c_1e^{-2t}cos \ t+c_2e^{-2t}sin \ t$
$$y'(t) = -c_1e^{-2t}(2cos\ t+sin \ t)+c_2e^{-2t}(cos \ t-2sin \ t)$$
Substitute the initial values in:
$$y(0) = c_1e^{0}cos \ 0+c_2e^{-0}sin \ 0 = 1$$
We get $c_1 = 1$.
$$y'(0) = -c_1e^{0}(2cos\ 0+sin \ 0)+c_2e^{0}(cos \ 0-2sin \ 0) = 0$$
$$-2c_1+c_2 = 0$$
$$-2+c_2 = 0$$
$$c_2 = 2$$
Thus the general solution of the equation is $y = e^{-2t}cos \ t+2e^{-2t}sin \ t$

5
##### Quiz-3 / TUT0702 Quiz 3
« on: October 11, 2019, 02:44:03 PM »
Question:
Verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
$$y" -2y'+y=0, y_1(t) = e^t, y_2(t) = te^t$$
Solution: $$y_1'(t) = e^t$$
$$y_1"(t) = e^t$$
Substitude:\begin{equation}\begin{split} y" -2y'+y &= e^t-2(e^t)+e^t\\
&=2e^t-2e^t\\
&= 0
\end{split}\end{equation}
Thus  y1 is a solution.

$$y_2'(t) = e^t+te^t$$
$$y_1"(t) = 2e^t+te^t$$
Substitude:\begin{split} y" -2y'+y &= 2e^t+te^t-2(e^t+te^t)+te^t\\
&=2e^t+te^t-2e^t-2te^t+te^t\\
&= 0
\end{split}
Thus y2is a solution.
\begin{equation}
\begin{split}
W &= y_1y_2'-y_2y_1'\\
&=e^t(e^t+te^t)-e^t(te^t)\\
&=e^{2t}+te^{2t}-te^{2t}\\
&=e^{2t}
\end{split}\end{equation}
Since W is not 0 for every value of t, y1 and y2 form a fundamental set of solution.

6
##### Quiz-2 / TUT0702 Quiz2
« on: October 04, 2019, 02:44:33 PM »
Question:
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$(\frac{sin y}{y} - 2e^{-x}sinx) + (\frac{cosy + 2e^{-x}cosx}{y})y' = 0, \ u = ye^x$$
Solution:
Check $$M_y = N_x$$
$$M_y = \frac{y cos y - sin y}{y^2}$$
$$N_x = \frac{2}{y}(-e^{-x}cos x -e^{-x}sinx)$$
$$M_y \neq N_x$$
So the equation is not exact.
Multiply both sides by u.
$$(e^x sin y - 2ysinx)+(cos ye^x +2 cosx)\frac{dy}{dx} = 0$$
$$M_y = e^xcosy - 2sinx$$
$$Nx = e^xcosy - 2sinx$$
$$M_y = N_x$$
So the equation is now exact.
So there exist a function $$\psi (x,y)$$ such that $$\psi_x = M, \psi_y = N.$$
$$\psi = \int_{}{}M dx = \int (e^x sin y - 2ysinx) dx = e^x siny +2 ycos x +h(y)$$
$$\psi_y = e^x cos y + 2 cos x +h'(y) = N = e^xcosy +2 cosx$$
$$So h'(y) = 0$$
$$H(y) = 0$$
$$\psi = e^x siny +2 ycos x$$
The solution is
$$e^x siny +2 ycos x = c$$

7
##### Quiz-1 / TUT0702 Quiz1
« on: September 27, 2019, 03:38:53 PM »
Question: \begin{equation*} \frac{dy}{dx} = \frac{x^2-3y^2}{2xy} \end{equation*}
Show that the given equation is homogeneous. Solve the differential equation.
The equation can be written as:
\begin{equation*} \frac{dy}{dx} = \frac{1-3 (\frac{y}{x})^2}{2(\frac{y}{x})}\end{equation*}
It can be expressed as a function of \begin{equation*}\frac{y}{x}\end{equation*} So the equation is homogeneous.
Let \begin{equation*} u = \frac{y}{x} \end{equation*}Then \begin{equation*} y = ux \end{equation*} and
\begin{equation*} \frac{dy}{dx} = u + x\frac{dv}{dx}\end{equation*}
Substitute, we get:
\begin{equation*} u + x\frac{dv}{dx} = \frac{1-3u^2}{2u}\end{equation*}
\begin{equation*} x\frac{dv}{dx} = \frac{1-3u^2}{2u} - u \end{equation*}
Observe the equation is separable.
\begin{equation*} \int_{}{} \frac{2u}{1-5u^2} du = \int_{}{} \frac{1}{x} dx \end{equation*}
\begin{equation*} -\frac{1}{5} ln|1-5u^2| = ln|x| + c_1\end{equation*}
\begin{equation*}- ln|1-5u^2| = 5 ln|x| + c\end{equation*}
Substitute back,
\begin{equation*}- ln|1-5\frac{y^2}{x^2}| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}- ln|\frac{x^2-5y^2}{x^2}| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|+ln|x^2| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|+2ln|x| = 5 ln|x| + c_2\end{equation*}
\begin{equation*}-ln|x^2-5y^2|-3ln|x| = c_2\end{equation*}
\begin{equation*}-ln|x^3(x^2-5y^2)| = c_3\end{equation*}
\begin{equation*}|x^3(x^2-5y^2) = c\end{equation*}
\begin{equation*}|x^3||x^2-5y^2| = c\end{equation*} where c is an arbitrary constant.

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