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### Topics - Fenglun Wu

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##### Quiz-4 / TUT0602 QUIZ 4
« on: October 18, 2019, 03:30:57 PM »
Find the general solution of the given differential equation.
$$y'' + 2y' + 2y = 0$$

Solve the equation $r^2+2r+2=0$ for $r$
$$r = \frac{-2 \pm \sqrt{2^2-4(1)(2)} }{2(1)}$$
$$= \frac{-2 \pm \sqrt{-4}}{2} = \frac {-2 \pm 2i}{2} = -1 \pm i$$

We know that $y = c_1e^{\lambda t}cos(\mu t) + c_2e^{\lambda t}sin(\mu t)$ and $\lambda = -1$ and $\mu = 1$

Therefore, the required general solution is
$$y = c_1e^{-t}cos(t) + c_2e^{-t}sin(t)$$

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##### Quiz-3 / TUT0602 QUIZ 3
« on: October 11, 2019, 02:00:13 PM »
Find the Wronskian of two solutions of the given differential equation without solving the equation.
$$cos(t)y'' + sin(t)y' - ty = 0$$
First, we divide both sides of the equation by $cos(t)$
$$y'' + \frac{sin(t)}{cos(t)}y' - \frac{t}{cos(t)}y = 0$$
Then, we have $p(t) = \frac{sin(t)}{cos(t)} = tan(t)$
Therefore, the Wroskian
$$W[y_1, y_2](t) = c \times exp(-\int p(t)dt)$$
$$= c \times exp(-\int tan(t)dt) = c \times exp(ln|cos(t)|) = c \times cos(t)$$

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##### Quiz-2 / TUT0602 QUIZ 2
« on: October 04, 2019, 01:36:38 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3 + x(1+y^2)y' = 0, ~~~~ \mu(x, y) = \frac{1}{xy^3}$$
First, let's show the given equation isn not exact.
Define $M(x, y) = x^2y^3, ~~ N(x, y) = x(1+y^2)$
$$M_y = \frac{\partial}{\partial y}[x^2y^3] = 3x^2y^2$$
$$N_x = \frac{\partial}{\partial x}[x(1+y^2)] = 1 + y^2$$
Since $M_y \neq N_x$, this implies that the given equation is not exact.

Next, show that the given equation multiplied by the integrating factor $\mu(x, y) = \frac{1}{xy^3}$ is exact.
The new equation becomes
$$\frac{1}{xy^3}x^2y^3 + \frac{1}{xy^3}x(1+y^2)y' = x + (y^{-3} + y^{-1})y' =0$$
Define $M'(x, y) = x, ~~ N'(x, y) = y^{-3} + y^{-1}$
$$M'_y = \frac{\partial}{\partial y}(x) = 0$$
$$N'_x = \frac{\partial}{\partial x}(y^{-3} + y^{-1}) = 0$$
Since $M'_y = N'_x$, this implies that the given equation is exact.

Thus, we know that there exists a function $\phi (x, y) = C$ which satisfies the give differential equation.
Also,
$$\frac{\partial \phi}{\partial x} = M'(x, y) = x$$
$$\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$$
Integrate $\frac{\partial \phi}{\partial x} = x$ with respect to $x$ we have
$$\phi (x, y) = \frac{1}{2}x^2 + h(y)$$
Take derivative on both sides with respect to $y$ we get
$$\frac{\partial \phi}{\partial y} = h'(y)$$
Since we know $\frac{\partial \phi}{\partial y} = N'(x, y) = y^{-3} + y^{-1}$
Then $$h'(y) = y^{-3} + y^{-1}$$
Integrate $h'(y)$ with respect to y we have
$$h(y) = -\frac{1}{2}y^{-2} + ln|y| + C$$
Therefore, we have
$$\phi (x, y) = \frac{1}{2}x^{2} - \frac{1}{2}y^{-2} + ln|y| = C$$
is the general solution to the given differential equation.

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##### Quiz-1 / TUT0602 QUIZ 1
« on: September 27, 2019, 02:00:02 PM »
$$\frac{dy}{dx} = \frac{x - e^{-x}}{y + e^y}$$
Rewrite the equation:

$$(e^{-x} - x) + (y + e^y)\frac{dy}{dx} = 0$$

The equation is of the form

$$M(x) +N(y)\frac{dy}{dx} = 0$$

Hence, it is separable.

$$\int (y + e^y) dy = \int (x - e^{-x}) dx$$
$$\frac{y^2}{2} + e^y = \frac{x^2}{2} + e^{-x} + c'$$

Let $$c = 2c'$$

Solution:
$$y^2 - x^2 + 2(e^y - e^{-x}) = c$$

where $$y + e^y \neq 0$$

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