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##### Quiz-5 / LEC 5101 QUIZ 5
« on: November 01, 2019, 02:00:17 PM »
Find the general solution of the given differential equation:

$y"+4y'+4y=t^{-2}e^{-2t},t>0.$

Let $y"+4y'+4y=0.$

$r^{2}+4r+4=0,r_{1}=-2=r_{2}.$

Thus,

$y_{c}(t)=C_{1}e^{-2t}+C_{2}te^{-2t}.$

$W=\begin{vmatrix}e^{-2t} & te^{-2t}\\ -2e^{-2t} & -2te^{-2t}+e^{-2t} \end{vmatrix}=-2te^{4t}+e^{4t}+2te^{4t}=e^{4t}.$

$W_{1}=\begin{vmatrix}0 & te^{-2t}\\ 1 & -2te^{-2t}+e^{-2t} \end{vmatrix}=-te^{-2t}.$

$W_{2}=\begin{vmatrix}e^{-2t} & 0\\ -2e^{-2t} & 1 \end{vmatrix}=e^{-2t}.$

Let $\mu_{1}=\int\frac{W_{1}(t)g(t)}{W(t)}dt=\int\frac{-te^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-ln(t).$

$\mu_{2}=\int\frac{W_{2}(t)g(t)}{W(t)}dt=\int\frac{e^{-2t}\times t^{-2}e^{-2t}}{e^{4t}}dt=-\frac{1}{t}.$

Therefore,

$y(t)=y_{1}(t)\mu_{1}+y_{2}(t)\mu_{2}+y_{c}(t)=-ln(t)e^{-2t}-e^{-2t}+C_{1}e^{-2t}+C_{2}te^{-2t}.$

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##### Quiz-2 / tut 5103 quiz 2
« on: October 04, 2019, 02:00:01 PM »
show the given equation is not exact but becomes exact when multiplied by the given integrating factor, then solve the equation.

x^(2) * y^(3) + x * (1+y^2) * y' = 0, u(x, y) = 1/(xy^3).

First, let's show the given DE x^(2) * y^(3) + x * (1+y^2)y' = 0 is not exact.

Define M = x^(2) * y^(3), N = x * (1 + y^2).

M_y = d/(dy) [x^(2) * y^(3)] = 3x^(2) * y^(2)

N_x = d/(dx) [x * (1 + y^2)] = 1 + y^2

Since 3x^(2)y^(2) != 1 + y^2, the given DE is not exact.

multiply each side of given DE by integrating factor u, we get

1/(xy^3) * x^(2) * y^(3) + 1/(xy^3) * x * (1 + y^2)y' = x + (y^(-3) + y^(-1)) * y' = 0

Let the new M = 1/(xy^3) x^(2) * y^(3), new N =  1/(xy^3) * x * (1 + y^2)

M_y = 0, N_x = 0, M_y = N_x and the new DE is exact.

there exist \phy (x, y) such that

\phy x = M, \phy y = N.

\phy x = M = 1/(x * y^3) * x^(2) * y^(3)

Integrating both side by x, we have

\phy = 1/2 * x^2 + h(y)

\phy y = h'(y) = N =  1/(xy^3) *  x * (1 + y^2)

h'(y) =  1/(xy^3) *  x * (1 + y^2)

Integrating both side by y, we have

h(y) = -0.5 * y^(-2) + ln|y| + C

Altogether, we have

\phy (x,y) = 0.5 * x^2 - 0.5 * y^(-2) + ln|y| + C

So our general solution is

0.5 * x^2 - 0.5 * y^(-2) + ln|y| = C.

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