# Recall Key Points of MA 1505:

## 1. Fundamental Theorem in Calculus: $(\int_{0}^{x} f(t)dt)^{'}=f(x)$

## 2. Integration by Parts: $\int f(x) dg(x)= f(x)g(x) - \int g(x) df(x)$

## 3. Derivatives and Integration: $( \sin x)^{'} = \cos x$ $(\cos x)^{'} = -\sin x$

Hyperbolic Sine: $\sinh x= \frac{e^{x}- e^{-x}}{2}$

Hyperbolic Cosine: $\cosh x=\frac{e^{x}+e^{-x}}{2}$ $(\sinh x)^{'}= \cosh x$ $(\cosh x)^{'}= \sinh x$

# MA 1506 Tutorials:

Ordinary Differential Equations:

## 1. Newton-Leibniz Formula $y=y(x)$ is a function with one variable $x$ with ordinary differential equation $y^{'}=f(x).$ The solution is $y=\int f(x) dx + C$ with some constant $C$.

## 2. Separable Equations $y=y(x)$ is a function with one variable $x$ with ordinary differential equation $N(y)y^{'}=M(x),$ where $N(y)$ is a function with one variable $y$ and $M(x)$ is a function with one variable $x.$

The solution is $\int M(x) dx = \int N(y) dy + C$ with some constant $C.$

## 3. One Order Ordinary Differential Equations $y=y(x)$ is a function with one variable $x$ with one order ordinary differential equation $y^{'}+P(x)y=Q(x).$ The integrating factor is $R(x)= exp ( \int P(x) dx).$ That means $d( R(x) y) = R(x) Q(x)$ and take the integration under x at the both sides, $R(x)y=\int R(x)Q(x) dx + C$ for some constant $C.$

Sometimes, we need to make some substitution as $z=y^{2}$ or $z=\frac{1}{y}$, since the following formulas: $2 y y^{'} = (y^{2})^{'},$ $-\frac{y^{'}}{y^{2}} = (\frac{1}{y})^{'}.$

If there is an initial condition $y(0)=A$ for the first order ordinary differential equation $y^{'} + P(x)y=Q(x),$ then we must make use of the initial condition to calculate the constant C after we solved the equation.