[140130] Review on Partial Fraction decomposition and First order linear ODE

Partial Fraction Decomposition

Basic Integrations

Example 1

$\displaystyle \int \dfrac{1}{x} dx=\ln(|x|)+C$

(Please note the absolute value sign and the constant added)

Example 2

$\displaystyle \int \dfrac{1}{x^2+1} dx=\arctan(x)+C$

Applying the substitution ${x=\tan(\theta)}$ , we have,

$\displaystyle \int \dfrac{1}{x^2+1} dx$

$\displaystyle =\int \cos^2(\theta) d(\tan(\theta))$

$\displaystyle =\int \cos^2(\theta) (\tan(\theta))'d(\theta)$

$\displaystyle =\int d(\theta)$

$\displaystyle =\theta+C$

$\displaystyle =\arctan(x)+C$

Example 3

$\displaystyle \int \dfrac{1}{x^2+2} dx=\dfrac{1}{\sqrt{2}}\arctan(\dfrac{1}{\sqrt{2}}x)+C$

Applying the substitution ${x=\sqrt{2}y}$ , we have,

$\displaystyle \int \dfrac{1}{x^2+2} dx$

$\displaystyle =\int \dfrac{\sqrt{2}dy}{2(y^2+1)}$

$\displaystyle =\dfrac{1}{\sqrt{2}}\int\dfrac{dy}{y^2+1}$

$\displaystyle =\dfrac{1}{\sqrt{2}}\arctan(y)+C$

$\displaystyle =\dfrac{1}{\sqrt{2}}\arctan(\dfrac{1}{\sqrt{2}}x)+C$

So far, we handle the integration of fractions which has denominators of degree less than or equal to 2. So a natural question comes out: what if this degree increases?

Example 4

$\displaystyle \int\dfrac{1}{(x+1)(x^2+2)}dx=?$

From Calculus II courses, we know by Partial Fraction Decomposition Theorem, which, loosely speaking, guarantees that any fraction can be decomposed to the sum of fractions, which has denominators of degree less than or equal to two. Thus,

$\displaystyle \dfrac{1}{(x+1)(x^2+2)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+2}=\dfrac{A(x^2+2)+(Bx+C)(x+1)}{(x+1)(x^2+2)}$

That is,

$\displaystyle A(x^2+2)+(Bx+C)(x+1)=1\ \ \ \ (*)$

or

$\displaystyle (A+B)x^2+(B+C)x+2A+C=1=0x^2+0x+1$

Therefore, we can solve out A,B,C by

$\displaystyle A+B=0$

$\displaystyle B+C=0$

$\displaystyle 2A+C=1$

which gives us ${A=-B=C=1/3}$ .
Thus,

$\displaystyle \int\dfrac{dx}{(x+1)(x^2+2)}=\dfrac{1}{3}\int\dfrac{dx}{x+1}-\dfrac{1}{3}\int\dfrac{xdx}{x^2+2}+\dfrac{1}{3}\int\dfrac{dx}{x^2+2}$

$\displaystyle =\dfrac{1}{3}\ln|x+1|-\dfrac{1}{6}\ln|x^2+2|+\dfrac{1}{3\sqrt{2}}\arctan(\dfrac{1}{\sqrt{2}}x)+C$

Remark 1 note (*) holds for any real x, so if we take ${x=-1}$ , we’ll get ${A=1/3}$ directly. To emphasize the convenience, we give another example,

$\displaystyle \dfrac{x^2+1}{x(x^2-1)}=\dfrac{A}{x}+\dfrac{B}{x+1}+\dfrac{C}{x-1}=\dfrac{A(x+1)(x-1)+Bx(x-1)+Cx(x+1)}{x(x+1)(x-1)}$

Take ${x=0}$ , we get ${A=-1}$ ;
Take ${x=1}$ , we get ${C=1}$ ;
Take ${x=-1}$ , we get ${B=1}$ ;
Another question is can we do this for all fractions? The answer is Yes by Partial Fraction Decomposition Theorem. Please check the link to consult the theorem and please do not forget the lower order terms in the decomposition, as the example following:

$\displaystyle \dfrac{x^2+1}{x(x-1)^3(x^2+2x+2)^2}=\dfrac{A}{x}+\dfrac{B}{x-1}+C\dfrac{1}{(x-1)^2}+\dfrac{D}{(x-1)^3}+\dfrac{Ex+F}{x^2+2x+2}+\dfrac{Gx+H}{(x^2+2x+2)^2}$

Differential Equations
For differential equations, we care about four issues about the solution: existence, uniqueness, stability (loosely speaking, whether the small variation of the initial condition will result in large variation on the solution) and long term behavior (the behavior when the variable goes to infinity)

Example 5

$\displaystyle y'=0$

has a solution

$\displaystyle y=C$

Example 6

$\displaystyle y'=1$

has a solution

$\displaystyle y=x+C$

Example 7

$\displaystyle y'+\alpha y=0 \ \ \ (\alpha\neq 0)$

We multiply ${e^{\alpha x}}$ on both side, (the reason we use this is that it is a solution of ${g'=\alpha g}$ , which will help us to convert the right hand side to be ${gy'+g'y}$ , the preferred form based on derivative rule of product)and it will help us convert the LHS to be a derivative of some function.

$\displaystyle \Longrightarrow e^{\alpha x}(y'+\alpha y)=0$

$\displaystyle \Longrightarrow (e^{\alpha x})'=0$

Integrating both side,

$\displaystyle \Longrightarrow e^{\alpha x}y=C$

$\displaystyle \Longrightarrow y=Ce^{-\alpha x}$

Example 8

$\displaystyle y'+\alpha y=1 \ \ \ (\alpha\neq 0)$

Applying the idea of example 7,

$\displaystyle \Longrightarrow e^{\alpha x}(y'+\alpha y)=e^{\alpha x}$

$\displaystyle \Longrightarrow (e^{\alpha x})'=e^{\alpha x}$

Integrating both side,

$\displaystyle \Longrightarrow e^{\alpha x}y=\dfrac{1}{\alpha}e^{\alpha x}+C$

$\displaystyle \Longrightarrow y=\dfrac{1}{\alpha}+Ce^{-\alpha x}$

Remark 2 Loosely speaking, the differential equations which have 0 on the RHS are called homogenous equations and the other cases are called inhomogenous equations. (Of course, you should put all terms related to y to the LHS before you make the judgement)

According to the examples above, we know how to handle the equations with constant coefficient, then what if the coefficient is related to x?

Example 9

$\displaystyle y'+f(x)y=0$

We apply the idea in Example 7 by multiplying ${e^{\int_0^x f(t)dt}}$ on both side. Then,

$\displaystyle e^{\int_0^x f(t)dt}(y'+f(x)y=0)$

$\displaystyle (e^{\int_0^x f(t)dt}y)'=0$

$\displaystyle y=Ce^{-\int_0^x f(t)dt}$

Finally, we handle a problem according to the above ideas,

Example 10

$\displaystyle y'+\frac{1}{x+1}y=\frac{1}{x^2+2}\ \ \ (x>0)$

Here, ${f(x)=1/(x+1)}$ , hence,

$\displaystyle e^{\int_0^x f(t)dt}=e^{\ln|x+1|}=|x+1|=x+1$

We multiply ${x+1}$ on both side, and get,

$\displaystyle (x+1)y'+y=\frac{1}{x^2+2}+\frac{x}{x^2+2}$

Integrating both side,

$\displaystyle (x+1)y=\frac{1}{\sqrt{2}}\arctan(\frac{1}{\sqrt{2}}x)+\frac{1}{2}\ln(x^2+2)+C$

which is,

$\displaystyle y=\dfrac{1}{\sqrt{2}(x+1)}\arctan(\frac{1}{\sqrt{2}}x)+\frac{1}{2(x+1)}\ln(x^2+2)+\frac{C}{x+1}$