Messages

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
$$   e^x siny +2 ycos x = c$$