Integration Techniques for Differential Equation

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First Order Differential Equation

1. Simple one. If the differential equation is like $\frac{\mathrm{d}x}{\mathrm{d}y} = Q(x)$, then solution would be $y=\int Q(x) \mathrm{d}x$.
2. Separation of variables. If the differential equation is like $\frac{\mathrm{d}x}{\mathrm{d}y} = f(x)g(y)$, then it could be re-write in $\frac{1}{f(x)}\mathrm{d}x = g(y) \mathrm{d}y$, then integrating each side and then the solution can be goten.
3. Integrating factor. If the differential equation is like $\frac{\mathrm{d}x}{\mathrm{d}y}+P(x)y=Q(x)y$, then it could become $$e^{\int P(x) \ \mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}y} + e^{\int P(x) \ \mathrm{d}x} \ P(x)y = e^{\int P(x) \ \mathrm{d}x} \ Q(x) $$ by multiplying $$e^{\int P(x) \ \mathrm{d}x}$$, so that
   $$e^{\int P(x) \ \mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}y} + e^{\int P(x) \ \mathrm{d}x} \ P(x)y = e^{\int P(x) \ \mathrm{d}x} \ Q(x) \ $$
   using product rule,
   $$ \frac{\mathrm{d}}{\mathrm{d}x} (e^{\int P(x) \ \mathrm{d}x} y) = e^{\int P(x) \ \mathrm{d}x} Q(x) $$

Second Order Differential Equations

General second order differential equation:
$$ a \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b \frac{\mathrm{d}y}{\mathrm{d}x} +cy = f(x)$$

1. Homogeneous equation: $$ a \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b \frac{\mathrm{d}y}{\mathrm{d}x} +cy = 0$$. We can solve it with substitution：let $y=e^{kx}$, then the equation becomes $$ak^2 e^{kx}+bke^{kx}+ce^{kx}=0$$ We know that $x \neq 0$, so that the equation becomes to $$ak^2+bk+c$$ which is a quadratic equation, and we can easily get the value of $k$, and the solutions are related to $k$.

• If the quadratic equation has 2 different real roots, then the solution of y is $y=Ae^{k_1t}+Be^{k_2t}$.
• If the quadratic equation has 2 same real roots, then the solution of y is $y=(A+Bx)e^{kt}$.
• If the quadratic equation has complex roots, such as $k=\alpha+i \beta$, then the solution of y is $y=(A \sin\beta x+B\cos\beta x)e^{\alpha x}$.

2. Non-homogeneous equation: the general solutions is the sum of the complementary function, that is the solution of the homogeneous version of the equation, and a particular integral. e.g.

• If $f(x)$ is a polynomial of degree $n$, guess $y = a_nx^{n}+a_{n-1}x^{n-1}+ \cdots +a_1x+a_0$.
• If $f(x)$ involves $e^{kx}$, guess $y=ce^kx$.
• If $f(x)$ involves $\sin kx$ or $\cos kx$, guess $y = A \sin kx + B \cos kx$.

Coupled Differential Equations

1. Simple way: Take the derivative of two known differential equations, and then use substitution elimination to calculate the general solution of one of them (usually a second order differential equation). Another general solution is obtained by using the known algebraic relations. It should be noted that the other general solution cannot be solved in previous way, because there is an algebraic relationship between the two solutions, and solving the second solution directly by substitution elimination will cause the relationship between the two to be lost, e.g.$$ \frac{\mathrm{d}x}{\mathrm{d}t} = 4x-2y \quad (1) \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = 3x-y \quad (2)$$ derivative of those two differential equation is that $$ \frac{\mathrm{d}^2x}{\mathrm{d}t^2} = 4 \frac{\mathrm{d}x}{\mathrm{d}t} - 2\frac{\mathrm{d}y}{\mathrm{d}t} \quad (3) \qquad \frac{\mathrm{d}^2y}{\mathrm{d}t^2} = 3 \frac{\mathrm{d}x}{\mathrm{d}t} - 2\frac{\mathrm{d}y}{\mathrm{d}t} \quad (4)$$ from (1), we can know that $$2y = 4x - \frac{\mathrm{d}x}{\mathrm{d}y}$$, so that $$y = 2x - \frac{1}{2}\frac{\mathrm{d}x}{\mathrm{d}y}$$ from (2), we know that $$\frac{\mathrm{d}y}{\mathrm{d}t} = 3x-y$$, and then bring them into (3), which is $$\begin{aligned} \frac{\mathrm{d}^2x}{\mathrm{d}t^2} &= 4 \frac{\mathrm{d}x}{\mathrm{d}t} - 2(3x-y) \\ &=4 \frac{\mathrm{d}x}{\mathrm{d}t} - 2(3x-2x - \frac{1}{2}\frac{\mathrm{d}x}{\mathrm{d}y}) \\ &= 4 \frac{\mathrm{d}x}{\mathrm{d}t} - 6x + 4x - \frac{\mathrm{d}x}{\mathrm{d}y} \\ &= 3 \frac{\mathrm{d}x}{\mathrm{d}t} -2x \end{aligned}$$ and now, the problem convert into a second order differential equation, and to solve this, just try $x=e^{kt}$, and then can easily get $x = Ae^{2t}+Be^t$, where $A$, $B$ are constant.
   then we use the algebraic relationship between x and y, that is $$\begin{aligned} y &=2x-\frac{1}{2}\frac{\mathrm{d}x}{\mathrm{d}t} \\ &=2(Ae^2t + Be^t) - \frac{1}{2} (2Ae^{2t} + Be^t) \\ &= Ae^{2t} + \frac{3}{2} Be^t \end{aligned}$$. So the solution is $$ x=Ae^{2t} + Be^t \qquad y=Ae^{2t}+\frac{3}{2}e^t$$, where $A$, $B$ are constant.

2. Another way: using matrix, that means, converted those two equation into a matrix $$Y = {x(t) \choose y(t)}, \qquad \dot{Y} = {\dot{x}(t) \choose \dot{y}(t)}$$, where $$\dot{Y} = MY$$, and the eigenvalues of $M$ are $\lambda_1, \lambda_2, \cdots \lambda_n$, and the corresponding eigenvectors are $v_1, v_2, \cdots, v_n$, where n is the number of eigenvalues of matrix M, and the solution of $$Y = \sum^{n}_{i=1} A_ie^tv_i$$. The advantage of this method is that it is more easy to calculate the solution for the multiple differential equations together, not only coupled. e.g. $$ \frac{\mathrm{d}x}{\mathrm{d}t} = 4x-2y \quad (1) \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = 3x-y \quad (2)$$, let $$Y = {x(t) \choose y(t)}, \qquad \dot{Y} = {\dot{x}(t) \choose \dot{y}(t)}$$, so that $$\dot{Y} = \begin{pmatrix} 4 & -2 \\ 3 &-1 \end{pmatrix}Y$$. then we can easily calculate the eigenvalues and the correponding eigenvectors, that is, for $\lambda = 2$, the eigenvector is $k{1 \choose 1}$, and for $\lambda = 1$, the eigenvector is $k{2 \choose 3}$. So the solution $$Y = Ae^{2t}{1 \choose 1} + Be^t{2 \choose 3}$$, where $A$, $B$ are constants, and if we expand it, we will get the same solution from the previous method.