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### Messages - Victor Ivrii

Pages:  2 3 ... 165
1
##### 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.

2
##### Chapter 7 / Re: Transforming a system of linear equations to a single higher-order equation
« on: November 05, 2019, 09:52:23 PM »
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.

3
##### 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 $y(0)=0$, $y'(0)=0$.

4
##### 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 $y(0)=0$, $y'(0)=0$.

5
##### 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 $y(0)=0$, $y'(0)=0$

6
##### Term Test 1 / Problem 4 (main)
« on: October 23, 2019, 06:24:24 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -6y' +10y =2e^{3x} +39\cos(x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

7
##### Term Test 1 / Problem 3 (afternoon)
« on: October 23, 2019, 06:13:03 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -5y'+6 y= 52\cos (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

8
##### Term Test 1 / Problem 3 (noon)
« on: October 23, 2019, 06:12:04 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -4y'+3 y= 96\sinh (x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

9
##### Term Test 1 / Problem 3 (morning)
« on: October 23, 2019, 06:11:02 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -6y'+8 y= 48\sinh (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Hint: $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$.

10
##### Term Test 1 / Problem 3 (main)
« on: October 23, 2019, 06:09:33 AM »
(a) Find the general solution for equation
Find the general solution for equation
\begin{equation*}
y'' -2y'-3y= 16\cosh (x).
\end{equation*}

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

11
##### Term Test 1 / Problem 2 (afternoon)
« on: October 23, 2019, 06:03:55 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
(2x+1)x y''+(2x+2)y'-2y=0.
\end{equation*}
(b) Check that $y_1(x)=x+1$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(-1)=1, y'(-1)=0}$.

12
##### Term Test 1 / Problem 2 (noon)
« on: October 23, 2019, 06:02:44 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
\bigl(x\cos(x)-\sin(x)\bigr)y''+x\sin(x)y'-\sin(x)y=0.
\end{equation*}
(b) Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(\pi)=\pi, y'(\pi)=0}$.

13
##### Term Test 1 / Problem 2 (morning)
« on: October 23, 2019, 06:00:35 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
x y''-(2x+1)y'+(x+1)y=0.
\end{equation*}
(b) Check that $y_1(x)=e^x$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(1)=0, y'(1)=e}$.

14
##### Term Test 1 / Problem 2 (main)
« on: October 23, 2019, 05:59:13 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE
\begin{equation*}
x^2 y'' -2xy' + (x^2+2)y=0
\end{equation*}
(b) Check that $y_1(x)=x\cos(x)$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(\frac{\pi}{2})=1, y'(\frac{\pi}{2})=0}$.

15
##### Term Test 1 / Problem 1 (afternoon)
« on: October 23, 2019, 05:57:42 AM »
(a) Find integrating factor and then a general solution of ODE
\begin{equation*}
-y^2\sin(xy) + \bigl(-xy \sin(xy)+2\cos(xy)+3y\bigr) y'=0
\end{equation*}

(b) Also, find a solution satisfying $y(\dfrac{\pi}{3})=1$.

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