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### Messages - Gavrilo Milanov Dzombeta

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##### Quiz 1 / Quiz 1 - Variant 2E - Problem 2
« on: January 29, 2021, 10:55:56 AM »
$$\text{Find the general solutions to the following equation: } u_{xxy}=x\cos(y)$$
\begin{gathered}
\nonumber
u_{xxy}=x\cos(y)\\
u_{xx}=x\sin(y) + f(x)\\
u_{x}=\frac{x^2}2\sin(y) + \tilde{f}(x) + h(y) ;\tilde{f}_x=f \\
u= \frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y);\tilde{\tilde{f}}_{xx}=f \\
\therefore u(x,y)=\frac{x^3}6\sin(y) + \tilde{\tilde{f}}(x) + xh(y) + g(y).\\
\text{Where $f(x),h(y),g(y)$ are arbitrary functions, and $\tilde{\tilde{f}}$ is twice differentiable.}
\end{gathered}

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##### Quiz 1 / Quiz 1 - Variant 2E - Problem 1
« on: January 29, 2021, 10:34:47 AM »
Consider first order equations and determine if they are linear homogeneous, linear inhomogeneous, or nonlinear ($u$ is an unknown function); for nonlinear equations, indicate if they are also semilinear or quasilinear:
$$u_t +u^2_x + u^2_y = 0.$$
$$\text{The equation is nonlinear homogeneous}$$

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##### Term Test 1 / Re: Problem 1 (main sitting)
« on: October 23, 2019, 06:36:48 AM »
$$M = y + 3y^2 e^{2x} \implies M_y = 1 + 6y e^{2x}$$
$$N = 1 + 2y e^{2x} \implies N_x = 4y e^{2x}$$
$$M_y \neq N_x \implies \text{ not exact}$$
$$\dfrac{M_y - N_x}{N} = 1$$
$$\mu = e^{\int 1 dx} = e^x$$
$$\left(e^x y + 3y^2 e^{3x}\right) + \left(e^x + 2y e^{3x}\right) y^\prime = 0$$
$$\tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\tilde{M_y} = e^x + 6y e^{3x}$$
$$\tilde{N} = e^x + 2y e^{3x}$$
$$\tilde{N_x} = e^x + 6y e^{3x}$$
$$\tilde{M_y} = \tilde{N_x} \implies \text{ exact equation}$$
$$\psi_x = \tilde{M} = e^x y + 3y^2 e^{3x}$$
$$\int \psi_x dx = \int \left(e^x y + 3y^2 e^{3x}\right) dx$$
$$\therefore \psi = e^x y + y^2 e^{3x} + h\left(y\right)$$
$$\psi_y = e^x + 2y e^{3x} + h^{\prime}\left(y\right) = \tilde{N}$$
$$\therefore h^{\prime}\left(y\right) = 0 \implies h\left(y\right) = c$$
$$\therefore \psi = e^x y + y^2 e^{3x} = c$$
$$y\left(0\right) = 1 \implies 1 + 1 = c$$
$$\therefore e^x y + y^2 e^{3x} = 2$$

There is no reason to post after this, unless you disagree with this solution. Especially no point to post attachments. V.I.
Instead of sequence single equations it would be better to use multiline environment like gather or gather* to avoid excessive vertical spacing
Code: [Select]
\begin{gather} EQUATION \\ EQUATION \\  .... \end{gather}

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##### Quiz-4 / Q4: TUT0102
« on: October 18, 2019, 02:00:43 PM »
$$\text{Solve the following given equation for t > 0.}$$
$$\text{No need to do an actual change of variables, just use the result of Problem 34, Section 3.3}$$
$$t^2 y^{\prime \prime} + 3 t y^{\prime} - 3y = 0$$

$$\text{characteristic equation: } r^2 - r + 3r -3 = 0$$
$$r^2 + 2r - 3 =0$$
$$\therefore r = \dfrac{-2 \pm \sqrt{16}}{2} \implies r_1 = 1 \text{ and } r_2 = - 3$$
$$y_1 = e^{x} \text{ and } y_2 = e^{- 3x}$$
$$\therefore y\left(x\right) = c_1 e^{x} + c_2 e^{- 3x}$$
$$x = \ln{t}$$
$$\therefore y\left(t\right) = c_1 e^{\ln{t}} + c_2 e^{- 3 \ln{t}}$$
$$\therefore y\left(t\right) = c_1 t + \dfrac{c_2}{t^3}$$

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##### Quiz-3 / Q3: TUT0102
« on: October 11, 2019, 02:00:06 PM »
$$\text{Verify that the functions } y_1 \text{ and } y_2 \text{ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?}$$
$$\left( 1- x \cot(x)\right) y^{\prime \prime} - x y^{\prime} + y = 0,$$
$$0 < x < \pi \text{ and } y_1 (x) = x \text{ and } y_2 (x) = \sin(x)$$

$$\text{Consider } y_1(x) = x$$
$$\therefore {y_1}^{\prime}(x) = 1 \text{ and } {y_1}^{\prime \prime}(x) = 0$$
$$(1 - x\cot(x))y^{\prime \prime} - xy^{\prime} + y = (1 - x\cot(x))(0) - x + x = 0$$
$$\therefore y_1(x) = x \text{ is a solution of the differential equation } (1 - x\cot(x))y^{\prime \prime} - x y^{\prime} + y = 0$$
$$\text{Consider } y_2(x) = \sin(x)$$
$$y^{\prime} = \cos(x) \text{ and } y^{\prime \prime} = \sin(x)$$
$$(1 - x\cot(x))y^{\prime \prime} - x y^{\prime} + y = (1 - x\cot(x))(-\sin(x)) - x(\cos(x)) + \sin(x) = 0$$
$$W(y_1, y_2)(t) = \begin{array}{|cc|} x & \sin(x) \\ 1 & \cos(x) \end{array} = x\cos(x) - \sin(x) \ne 0 \text{ for } 0 < x < \pi$$
$$\therefore y_1, y_2 \text{ form a fundamental set of solutions}$$

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##### Quiz-2 / Q2: TUT0102
« on: October 04, 2019, 02:00:43 PM »
$$\text{Find an integrating factor and solve the following equation: }$$
$$e^{x} + \left(e^{x} \cot(y) + 2y \csc(y)\right)y^\prime = 0 \tag{1}$$

$$\text{Let } M = e^{x} \text{ and } N = e^{x} \cot(y) + 2y \csc(y)$$
$$\text{Then } M_y = 0 \text{ and } N_x = e^{x} \cot(y)$$
$$M_y \neq N_x \implies \text{The equation is not exact.}$$
$$\dfrac{N_x - M_y}{M} = \dfrac{e^{x}\cot(y)}{e^{x}} = \cot(y)$$
$$\mu = e^{\int{\cot(y)dy}} = e^{\int \frac{\cos(y)}{\sin(y)}dy}$$
$$\text{The integrating factor is: } \mu = \sin(y)$$
$$\text{Multiply equation 1 by \mu.}$$
$${e^{x}}\sin(y) + \left({e^{x}}\cos(y) + 2y\right)y^\prime = 0 \tag{2}$$
$$\text{ Let } \tilde{M} = e^{x}\sin(y) \text{ and } \tilde{N} = {e^{x}}\cos(y) + 2y$$
$$\tilde{M}_y = {e^{x}}\cos(y) \text{ and } \tilde{N}_x = {e^{x}}\cos(y)$$
$$\tilde{M}_y = \tilde{N}_x \implies \text{ Equation 2 is exact.}$$
$$\psi_x = M \tag{3}$$
$$\text{Integrating equation 3 with respect to x},$$
$$\int{\psi_x dx} = \int{e^{x} \sin(y) dx}$$
$$\therefore \psi = e^{x} \sin(y) + h(y) \tag{4}$$
$$\psi_y = e^{x} \cos(y) + h^\prime(y)$$
$$\psi_y = N$$
$$\therefore e^{x} \cos(y) + h^\prime(y) = e^{x}\cos(y) + 2y$$
$$\int{h^\prime(y) dy} = \int{2y dy} \implies h(y) = y^{2}$$
$$\therefore e^{x}\sin(y) + y^{2} = c$$

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##### Quiz-1 / Q1: TUT0102
« on: September 27, 2019, 02:01:46 PM »
$$\text{Find the general solution to the given equation by variation of parameter }$$
$$y' + \dfrac{1}{t}y = 3\cos(2t) ,t\gt 0$$
$$\text{Make the equation homogeneous}$$
$$y' + \dfrac{1}{t}y = 0$$
$$\dfrac{dy}{dt} = - \dfrac{y}{t}$$
$$\dfrac{dy}{y} = - \dfrac{dt}{t}$$
$$\int \dfrac{dy}{y} = - \int \dfrac{dt}{t}$$
$$ln(y) = - ln(t) + c$$
$$y = e^{-ln(t)} e^{c}$$
$$\text{let } A = e^{c}$$
$$y = \dfrac{A}{t} \implies A(t) = yt$$
$$y = \dfrac{A(t)}{t}$$
$$y' + \dfrac{1}{t}y = 3\cos(2t)$$
$$\left[\dfrac{A(t)}{t}\right]^{'} + \dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$\dfrac{tA'(t) - A(t)}{t^2} +\dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$\dfrac{A'(t)}{t} = 3\cos(2t)$$
$$A'(t) = 3t\cos(2t)$$
$$\int A'(t) dt = \int 3t\cos(2t) dt$$
$$\text{Integration by parts}$$
$$A(t) = \dfrac{3t\sin(2t)}{2} + \dfrac{3\cos(2t)}{4} + C$$
$$y = \dfrac{3\cos(2t)}{4t} + \dfrac{3\sin(2t)}{2t} + \dfrac{C}{t}$$

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