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08/23/11 - Classifying DEs, Direction Fields and Numerical Solutions, and Separation of Variables
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  | Last time we found that we could simply solve ODEs of the form y'(x) = f(x) by integration.
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  | Integrate right side by parts:








 Integral of left-hand side is y(x) - y(0):



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  | Dependent and Independent Variables and ODEs and PDEs
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  | When we write y = f(x) or y = g(t). x or t is called the independent variable (meaning we can choose its value freely), and y is called the dependent variable (meaning its value is determined by the x or t value we choose).
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  | If a function has only one independent variable (usually x or t), then any equation involving the independent variable (say t) and the dependent variable (y) and its derivatives (y', y'',...) is called an ordinary differential equation (ODE). That is: y'(t) = f(y,t) or y''(t) = g(y,y',t).
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  | If a function has more than one independent variable (usually y = f(x,t) or z = g(x,y,t)), then any equation involving the independent variables and the dependent variable and its derivatives with respect to any of its independent variables is called a partial differential equation (PDE).
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  | The order of an ODE is given by the degree of its highest derivative.
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  | A first-order ODE has a first derivative of its dependent variable and may involve its dependent variable, the independent variable and some constants:

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  | A second-order ODE has a second derivative of its dependent variable and may involve its first derivative, its dependent variable, the independent variable and some constants:

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  | A third-order ODE has a third derivative of its dependent variable and may involve …?
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  | First-order ODEs can always be put into the form: f(t,y,y') = 0
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  | Frequently we put them in a form where we solve for the highest derivative (called normal form).
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  | Second-order: y'' = h(t,y,y')
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  | n-th order: y(n) = p(t,y,y',y'',...,y(n-1))
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  | Put this ODE into normal form: t+4yy' = 0
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  | Find a first-order ODE for which y = 3te-t is a solution.
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  | Initial-value problems (IVPs)
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  | Given that y = -1/(t-C) is a solution to y' = y2, find the particular solution satisfying y(0) = 1.
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  | y' = y2 has solution y = -1/(t-C). Interval of existence?
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  | y' = x + y has solution y = -1 - x + Cex. Interval of existence?
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  | s' = r1/2 has solution s = (2/3) r3/2 + C. Interval of existence?
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  | Geometric meaning of solutions (direction field)
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  | Draw the direction field for y' = -y + x for the region -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2.
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  | There are three types of ODEs that can be solved by separation of variables. ODEs in normal form which are the product a function of the dependent variable and a function of the independent variable
 or the quotient of two such functions:
 or
 are separable. That means they can be written in a form where one side of the equation depends only on the dependent variable and the other side depends only on the independent variable, and they can be integrated independently.
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  | Read sections 2.1 and 2.2 of the text. Do exercises p. 25 (3,6,7,8,13,18,24,25,31,38) and p. 35 (7,10,18,21,23,27,30,32,35,42).
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