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Messages - Victor Ivrii

Pages: 1 [2] 3 4 ... 166
16
Term Test 2 / Problem 3 (noon)
« on: November 19, 2019, 04:22:43 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} 1 &2\\
1 &0\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

(b) Find the general solution
$$
\mathbf{x}'=\begin{pmatrix} 1 &2\\
1 &0\end{pmatrix}\mathbf{x}+
\begin{pmatrix} 0 \\[1pt]
\dfrac{6 e^{3t }}{e^{2t}+1}\end{pmatrix}.$$

17
Term Test 2 / Problem 3 (morning)
« on: November 19, 2019, 04:21:57 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} -2 &1\\
-1 &0\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

(b) Find the general solution
$$
\mathbf{x}'=\begin{pmatrix} -2 &1\\
-1 &0\end{pmatrix}\mathbf{x}+
\begin{pmatrix} 0  \\
\dfrac{e^{-t}} {t^2+1} \end{pmatrix}.
$$

18
Term Test 2 / Problem 3 (main sitting)
« on: November 19, 2019, 04:21:01 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} 1 &1\\
-2 &4\end{pmatrix}\mathbf{x}
$$
classify fixed point $(0,0)$ and sketch trajectories.

(b) Find the general solution
$$
\mathbf{x}'=\begin{pmatrix} 1 &1\\
-2 &4\end{pmatrix}\mathbf{x}+
\begin{pmatrix} \dfrac{e^{4t }}{e^{2t}+1} \\
0\end{pmatrix}.
$$

19
Term Test 2 / Problem 2 (noon)
« on: November 19, 2019, 04:19:53 AM »
Consider equation
\begin{equation}
y'''+y''+4y'+4y=-24e^{-2t}.
\label{2-1}
\end{equation}
(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).

20
Term Test 2 / Problem 2 (morning)
« on: November 19, 2019, 04:18:46 AM »
Consider equation
\begin{equation}
y'''-2y''+4y'-8y=15\cos (t).
\label{2-1}
\end{equation}
(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).

21
Term Test 2 / Problem 2 (main sitting)
« on: November 19, 2019, 04:17:26 AM »
Consider equation
\begin{equation}
y'''+4y''+y'-6y=24e^{t}.
\label{2-1}
\end{equation}
(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).

22
Term Test 2 / Problem 1 (noon)
« on: November 19, 2019, 04:15:32 AM »
(a) Find the general solution of
$$
y''-3y'+2y=\frac{e^{3t}}{e^{2t}+1}.
$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$

23
Term Test 2 / Problem 1 (morning)
« on: November 19, 2019, 04:14:47 AM »
(a) Find the general solution of
$$
y''-y=\frac{12}{e^{t}+1}.
$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$

24
Term Test 2 / Problem 1 (main sitting)
« on: November 19, 2019, 04:13:51 AM »
(a) Find the general solution of
$$
y''+4y=\frac{1}{\cos^2(t)},\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.
$$

(b) Find solution satisfying
$$y(0)=y'(0)=0.$$

25
Chapter 4 / Re: 4.2
« on: November 15, 2019, 02:42:45 PM »
Yes, because for equations given they could be found easily

26
Chapter 7 / Re: Finding linear independence
« on: November 11, 2019, 06:58:04 AM »
Those are not just vectors, but vector-valued functions and you need to check that for constant coefficients  their linear combination is identically $0$ if and only if these coefficients are $0$.
Try first to look at components of this vector-function.

27
Nothing to do with the determinant. In dimension 2 you can reduce if the matrix is not diagonal--obvious. What about the diagonal case? It is also possible unless a matrix is scalar, i.e. proportional to identity. To do this reduction, we make first a linear transform, so that after it the matrix is not  diagonal anymore, and then reduce.

General criteria: System with constant coefficients could be reduced to a single equation iff each eigenspace is $1$-dimensional. To understand why we need to consider solutions to a homogeneous equation and to a homogeneous system.

For an equation one of the solutions is $t^{m-1} e^{kt}$ where $k$ is characteristic root, and $m$ is a multiplicity of $k$.

For system all solutions are in the form  $P_{s-1}(t)e^{kt}$ where $P_{s-1}(t)$ are polynomials of degree $\le s-1$ with vector-coefficients and $s$ is the maximal dimension of the corresponding Jordan cells.

Therefore reduction can be done iff $s=m$ which means that for each eigenvalue $k$ there is just one cell, which in turn means, that there is only one linearly independent eigenvector.

Remark: If $s_1,...,s_j$ are dimensions of all cells, corresponding to $k$, then their sum $=m$ where $m$ is a multiplicity of $k$ as a root of characteristic equation, and also the dimension of the root subspace, and $j$ is a dimension of the corresponding eigenspace.




However, it is not important: we solve systems without reducing them to  single equations.

28
Term Test 1 / Problem 4 (afternoon)
« on: October 23, 2019, 06:27:03 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' +2y' +17 y =40 e^{x} +130\sin(4x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

29
Term Test 1 / Problem 4 (noon)
« on: October 23, 2019, 06:26:03 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -8y' +25y =18e^{4x} +104\cos(3x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

30
Term Test 1 / Problem 4 (morning)
« on: October 23, 2019, 06:25:04 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -6y' +25y =16e^{3x} +102\sin(x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$

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