Find the solution of the equation
\[ \frac{dy}{dx} - P(x)y = e^{2x}, \quad y(0) = 1, \]
where \[P(x) = \begin{cases} 1 & x \in [0,1), \\ 0 & x \notin [0,1). \end{cases}\]
[Solution:] Consider the homogeneous equation
\[ \frac{dy}{dx} - P(x)y = 0, \]
by separating the variables we have \(\frac{dy}{y} = P(x)dx\), namely \(\ln|y| = \int_0^x P(t)dt + C_1\), and we get
\[ y = Ce^{\int_0^x P(t)dt} = \begin{cases} C & x < 0 \\ Ce^x & x \in [0,1) \\ Ce & x \geq 1. \end{cases} \]
Now by variation of parameters we suppose \(u = u(x)\), and \(y = u(x)e^{\int_0^x P(t)dt}\) is the solution, hence
\[ u'(x) = e^{2x - \int_0^x P(t)dt} = \begin{cases} e^{2x} & x < 0, \\ e^x & x \in [0,1), \\ e^{2x-1} & x \geq 1. \end{cases} \]
(this is because :
The differential equation is given by: \[ y' + p(x)y = f(x). \]
Integrating factor is: \[ e^{\int p(x)dx}. \]
Multiplying through by the integrating factor: \[ \frac{d}{dx} \left[ e^{\int p(x)dx} y \right] = f(x) e^{\int p(x)dx}. \]
Solving for \(y\):
\[ y e^{\int p(x)dx} = \int f(x) e^{\int p(x)dx} dx + C. \]
Thus, the general solution is: \[ y = Ce^{-\int p(x)dx} + e^{-\int p(x)dx} \int f(x) e^{\int p(x)dx} dx, \] where \(y_c\) represents the complementary solution \(Ce^{-\int p(x)dx}\) and \(y_p\) represents the particular solution \(e^{-\int p(x)dx} \int f(x) e^{\int p(x)dx}\).
)
and \(C(0) = 1\) since \(y(0) = 1\).
We see that
\[ u(x) = u(0) + \int_0^x u'(t)dt = \begin{cases} \frac{e^{2x} + 1}{2} & x < 0 \\ e^x & x \in [0,1) \\ \frac{e^{2x-1} + e}{2} & x \geq 1. \end{cases} \]
Finally we get
\[ y = \begin{cases} \frac{e^{2x} + 1}{2} & x < 0 \\ e^{2x} & x \in [0,1) \\ \frac{e^{2x} + e^2}{2} & x \geq 1. \end{cases} \]
plot of 3 cases damped motion
p31 可以震荡在0附近吗 pass x at most 1 time？
undamped is similar to damped
wronskian 是判断函数之间是否线性无关的一个方法 用求导
而wronskian for linear system是判断线性系统的解是否线性无关的一个方法 不用求导
这个3用背锅吗
Differential equations are the heart of analysis, which is the natural goal of elementary calculus and the most important part of mathematics for understanding the physical sciences.
We can use functions in a orthogonal set,
\[\left\{ 1, \cos\left(\frac{\pi}{p}x\right), \cos\left(\frac{2\pi}{p}x\right), \cos\left(\frac{3\pi}{p}x\right), \ldots, \sin\left(\frac{\pi}{p}x\right), \sin\left(\frac{2\pi}{p}x\right), \sin\left(\frac{3\pi}{p}x\right), \ldots \right\}\]
(similar to a orthognal basis in linear algebra, to obtain the Fourier series). Specifically, we start with the orthogonal set {1, sin, cos, …}. By projecting the function onto these basis functions or computing the inner product three times—(f, 1), (f, cos), and (f, sin)—we can derive the Fourier series representation of the function.
f, and f’ are piecewise continuous on [-p,p], then for every x $$ [-p,p], the Fourier series of f(x) converges to f(x) at every point x in [-p,p] where f is continuous. If f has a jump discontinuity at x, the Fourier series converges to the average of the left-hand and right-hand limits of f at x.
\(y'+ky = f(x)\) is just \((D+k)y=f(x)\)–> y=1/(D+k)f(x)
\(e^{kx} (y'+ky) =e^{kx} f(x)\)
\(ye^{kx} = \int e^{kx} f(x) dx\)
y= \(\frac{1}{e^{kx}} \int e^{kx} f(x) dx\) = \(e^{-kx} \int^m e^{km} f(m) dm\)–convolution
So the solution of y = \(\frac{1}{e^{kx}} \int e^{kx} f(x) dx\) is the same as the solution of y=1/(D+k)f(x)
eg. \(1/(D-2) (xe^x) = e^{-2x} \int e^{x} e^{2x} x dx = e^{-2x} \int e^{3x} x dx = e^{-2x} (e^{3x} (1/3)x - 1/9 e^{3x}) = 1/3 x e^x - 1/9 e^x\)
result <- 0.927 /6
tan(5.356)[1] -1.333028atan(5.356)[1] 1.386215tan(-0.9271752)[1] -1.3330.8*cos(8 *pi/12)[1] -0.40.8*cos(8 *pi/8)[1] -0.80.8*cos(8 *pi/6)[1] -0.40.8*cos(8 *pi/4)[1] 0.80.8*cos(8 * (9*pi/32))[1] 0.56568540.8*cos(8 * 3*pi/16)[1] -1.469576e-16-6.4*sin(8*3*pi/16)[1] 6.4….
We can differentiate \(e^{tA}\) with respect to \(t\) and obtain
\[ \begin{align*} \frac{d}{dt} e^{tA} &= A + tA^2 + \frac{t^2}{2!}A^3 + \frac{t^3}{3!}A^4 + \cdots \\ &= A \left( 1 + tA + \frac{t^2}{2!}A^2 + \frac{t^3}{3!}A^3 + \cdots \right) \\ &= Ae^{tA} \end{align*} \] as one would expect from differentiating the exponential function. We can therefore formally write the solution of
\[ \dot{x} = Ax \qquad x(0) = e^{tA}x(0) \]
If the matrix \(A\) is diagonalizable such that \(A = S\Lambda S^{-1}\), then observe that
\[ \begin{align*} e^{tA} &= e^{tS\Lambda S^{-1}} \\ &= I + tS\Lambda S^{-1} + \frac{t^2(S\Lambda S^{-1})^2}{2!} + \frac{t^3(S\Lambda S^{-1})^3}{3!} + \cdots \\ &= I + tS\Lambda S^{-1} + \frac{t^2 S\Lambda^2 S^{-1}}{2!} + \frac{t^3 S\Lambda^3 S^{-1}}{3!} + \cdots \\ &= S \left( I + t\Lambda + \frac{t^2\Lambda^2}{2!} + \frac{t^3\Lambda^3}{3!} + \cdots \right) S^{-1} \\ &= Se^{t\Lambda}S^{-1} \end{align*} \]
If \(\Lambda\) is a diagonal matrix with diagonal entries \(\lambda_1, \lambda_2,\) etc., then the matrix exponential \(e^{t\Lambda}\) is also a diagonal matrix with diagonal entries given by \(e^{\lambda_1 t}, e^{\lambda_2 t},\) etc.
We can now use the matrix exponential to solve a system of linear differential equations.
\(\textbf{Example: Solve the previous example}\)
\[ \mathbf{X}' = \begin{pmatrix} 4 & 4 & 4 \\ 4 & 4 & 4 \\ -8 & -8 & -8 \end{pmatrix} \mathbf{X} \]
We could use the traditional method of finding eigenvalues and eigenvectors, but more quicker is to use the matrix exponential since
\[ A^2=\begin{pmatrix} 4 & 4 & 4 \\ 4 & 4 & 4 \\ -8 & -8 & -8 \end{pmatrix}\begin{pmatrix} 4 & 4 & 4 \\ 4 & 4 & 4 \\ -8 & -8 & -8 \end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]
So we have \(e^{At}=I+At\)
Therefore, the solution of this system is \(e^{At}\vec C =e^{At}(C_1,C_2,C_3)\), which is
\[ X(t) = \begin{pmatrix} C_1 + 4t(C_1 + C_2 + C_3) \\ C_2 + 4t(C_1 + C_2 + C_3) \\ C_3 - 8t(C_1 + C_2 + C_3) \end{pmatrix} \]
\(\textbf{Example: Solve the previous example}\)
\[ \frac{d}{dt} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \]
by matrix exponentiation.
We know that
\[ \Lambda = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}, \quad S = \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix}, \quad S^{-1} = \frac{1}{4} \begin{pmatrix} -2 & -1 \\ -2 & 1 \end{pmatrix}. \]
The solution to the system of differential equations is then given by
$$ \[\begin{align*} x(t) &= e^{tA}x(0) \\ &= S^{A}S^{-1}x(0) \\ &= \frac{1}{4} \begin{pmatrix} 1 & 1 \\ 2 & -2 \end{pmatrix} \begin{pmatrix} e^{3t} & 0 \\ 0 & e^{-t} \end{pmatrix} \begin{pmatrix} -2 & -1 \\ -2 & 1 \end{pmatrix} \times (0). \end{align*}\]
$$
Successive matrix multiplication results in \[ \begin{align*} x_1(t) &= \frac{1}{4}(2x_1(0) + x_2(0))e^{3t} + \frac{1}{4}(2x_1(0) - x_2(0))e^{-t}, \\ x_2(t) &= \frac{1}{2}(2x_1(0) + x_2(0))e^{3t} - \frac{1}{2}(2x_1(0) - x_2(0))e^{-t}. \end{align*} \]
\(x^{(2)}\)= x’’ = \(\frac {d^2 x}{dy^2}\)–this notation of the second order’s is actually share the same idea of “division”
$a_n(x) + … +a_1(x) +a_0(x)y =g(x) $
(y-x)dx+4xdy=0
An funtion defined on an interval I and possessing at least n derivatives that are continuous on I, which when substituted into an nth-order ordinary differential equation reduces the equation to an identity, is said to be a solution of the equation on the interval.
We cannot think solution of an ordinary differential equation without simultaneously thinking interval.
System:
\[\cases{^{{\frac{dx}{dt}}=f(t,x,y)} _{\frac{dy}{dt}=g(t,x,y)}}\] Solution: x =\(\Phi_1(t)\), y=\(\Phi_2(t)\) defined on a common interval I that satisfy each equation of the system on this interval.
On some interval I containing \(x_0\), the problems of solving an nth-order DE subject to n side conditions (initial conditions (IC)) specified at \(x_0\)
Does a solution of the problem exist? If a solution exists, is it unique?
Suppose y(x) represents a solution of the first order initial- value problem. The following three sets on the real x-axis may not be the same:
the domain of the function y(x), D
the interval I over which the solution y(x) is defined or exists,
and the interval \(I_0\) of existence and uniqueness.
\(I_0 \subseteq I \subseteq D\)
for y(x): \(\frac{dy}{dx} = f(y)\)
Critical Points (equilibrium point/ stationary point)–f(y)=0
\(\frac{dp}{dt} = p(a-bp)\)
let f(p)=p(a-bp)=0 and catogorize the x-axis based on it, discussing the different situation intuitively
If y(x) is a solution of an autonomous differential equation dy ∕dx = f(y), then y1(x) = y(x − k) (y of (x-k)), k a constant, is also a solution.
\(\frac {dy}{dx}=g(x)h(y)\)
\(\frac {dy}{h(y)}=g(x)dx\). If r is a zero of the function h(y), substituting y=r into dy/dx=g(x)h(y) makes both sides zero (y = r is a constant solution of the differential equation). As a consequence, y = r might not show up in the family of solutions that are obtained after integration and simplification. Such a solution is called a singular solution.
\(a_1(x)\frac{dy}{dx}+a_0(x)y=g(x)\) is said to be a linear equation in the variable y.
\(\frac{dy}{dx}+P(x)y=f(x)\)
idea: want to written into \(\frac{d}{dx}()=f(x)\) then we assume after multiplying a factor \(\mu(x)\) we can write \(\frac{d}{dx}(\mu(x)y(x))=\mu(x)f(x)\)
we get \(\mu(x) = ce^{\int{p(x)dx}{}}\).
let c=1 we get the integrating factor
z=f(x,y)=c
DO Not Remember–understand it instead!
just want to transform the equation to a 1st order differential equation to solve it using integrating factor method.
Just follow the idea in \(y''+my'+ny=0\), when facing the linear system \(\cases{{\vec {x'} = A\vec x}\\{ay'+by=0}}\), we also assume that \(\vec {x'} = \vec ke^{\lambda t}\) and substitute it into \(\vec {x'} = Ax\)
When we solve \(ay''+by'+cy=0\) if m is a double root, then \(y_1=e^{mx}\), \(y_2=xe^{mx}\)
Naturally, here we assume \(\vec x_2 =\vec k t e^{\lambda t}\), substitute \(\vec x_2\) into the system \(\vec {x'} = Ax\);
LHS = \(\vec {x_2'} = \vec k(e^{\lambda t }+\lambda te^{\lambda t})=A\vec x_2\)=\(A\vec k t e^{\lambda t}\) =RHS
Then we get \(\vec k = \vec 0\).
So our assumption is not true.
There is an extra term