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Messages - wenlinwang

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a) $y^{'''} + 0y^{''} - 7y' + 6y =100e^{-3t}$

 p(t) = 0
 
 $$W = ce^{-\int p(t)dt} = ce^{c_1} = C$$

b) $r^3 -7r + 6 = 0$
$(r-1)(r-2)(r-3) = 0$
so r =1 r=2 r=-3
so $y(t) =c_1e^t + c_2e^{2t} + c_3e^{-3t}$

 \begin{equation*}
  W =  \begin{bmatrix}
    e^t & e^{2t} & e^{-3t}\\
    e^t & 2e^{2t} & -3e^{-3t}\\
    e^t & 4e^{2t} & 9e^{-3t}
    \end{bmatrix} = 20
\end{equation*}
so W = C = 20

c) $y_p(t) = Ae^{-3t}$ same with home part
so $y_p(t) = Ate^{-3t}$
$y_p'(t) = Ae^{-3t} - 3Ate^{-3t}$

$y_p''(t) = 9Ate^{-3t} - 6Ae^{-3t}$

$y_p'''(t) = 27Ae^{-3t} - 27Ate^{-3t}$

$y^{'''} + 0y^{''} - 7y' + 6y =100e^{-3t}$

$27Ae^{-3t} - 27Ate^{-3t} -7Ae^{-3t} + 21Ae^{-3t} + 6Ate^{-3t} = 100e^{-3t}$
so A = 5
so $y_p(t) = 5te^{-3t}$
so $y(t) = y_c(t) + y_p(t)$
$y(t) = c_1e^t +c_2e^{2t} + c_3e^{-3t} + 5te^{-3t}$

2
Find the general solution of the system of ODEs
$x_t' = x + y + \frac{e^t}{\cos (t)}$
$y_t' = -x + y + \frac{e^t}{\sin(t)}$

3
MAT244--Lectures & Home Assignments / Re: Integrating Factor problem
« on: November 27, 2018, 05:10:46 PM »
$$M_y: 2xcos(y) -2xysin(y)-2ycos(x)\\
Nx:4xcos(y)-2xysin(y)-3ycos(x)\\
M_y \neq N_x\\

R = \frac{M_y -N_x}{M} =\frac {-2xcos(y) + ycos(x)}{2xycos(y) - y^2cos(x)}= \frac {-(2xcos(y)-ycos(x)}{2xycos(y)-y^2cos(x)}=\frac{-1}{y}\\

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

both side times y
$$
(2xy^2cos(y)-y^3cos(x))dx + (2x^2ycos(y) -y^2x^2sin(y)-3y^2sin(x)-5y^4)dy = 0

\phi_{(x,y)} , \phi_x = M, \phi_y = N \\

\phi = \int{M}dx = \int{2xy^2cos(y) - y^3cos(x)}dx\\
      = x^2y^2cos(y) - y^3sin(x) + h(y)
\phi_y = 2x^2ycos(y) - x^2y^2sin(y) -3y^2sin(x) + h'(y)= N\\
=2x^2ycos(y)-y^2x^2sin(y)-3y^2sin(x)-5y^4\\

 h'(y) = -5y^4\\
h(y) = -y^5\\
\phi = x^2y^2cos(y) - y^3sin(x)-y^5 = C
$$

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