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Topics - Gavrilo Milanov Dzombeta

Pages: [1]
1
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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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 »
$$\text{Find the general solution to the given equation by variation of parameter } $$
$$ y' + \dfrac{1}{t}y = 3\cos(2t) ,t\gt 0$$
$$\text{Make the equation homogeneous} $$
$$ y' + \dfrac{1}{t}y = 0 $$
$$\dfrac{dy}{dt} = - \dfrac{y}{t}$$
$$\dfrac{dy}{y} = - \dfrac{dt}{t} $$
$$\int \dfrac{dy}{y} = - \int \dfrac{dt}{t} $$
$$ ln(y) = - ln(t) + c $$
$$ y = e^{-ln(t)} e^{c}$$
$$ \text{let } A = e^{c} $$
$$ y = \dfrac{A}{t} \implies A(t) = yt $$
$$ y = \dfrac{A(t)}{t} $$
$$ y' + \dfrac{1}{t}y = 3\cos(2t) $$
$$ \left[\dfrac{A(t)}{t}\right]^{'} + \dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$ \dfrac{tA'(t) - A(t)}{t^2} +\dfrac{1}{t^2}\left[A(t)\right] = 3\cos(2t)$$
$$ \dfrac{A'(t)}{t} = 3\cos(2t) $$
$$ A'(t) = 3t\cos(2t) $$
$$ \int A'(t) dt = \int 3t\cos(2t) dt$$
$$\text{Integration by parts} $$
$$ A(t) = \dfrac{3t\sin(2t)}{2} + \dfrac{3\cos(2t)}{4} + C $$
$$ y = \dfrac{3\cos(2t)}{4t} + \dfrac{3\sin(2t)}{2t} + \dfrac{C}{t} $$

Pages: [1]