chapter2.page05

# chapter2.page05 - y = x 2 3 Solution The solution is y x =...

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2. SEPARABLE EQUATIONS 5 (c) R x - 3 ( x 2 +3)( x +5) dx (d) R x tan( x 2 + 1) dx 2. Separable equations If we let x be the independent variable and y be the dependent variable, a separable ODE is given by, y 0 = f ( x ) g ( y ) where both f and g are given one variable functions and y 0 = dy dx is the derivative of y ( x ) with respect to x and y ( x ) is the unknown function we want to ﬁnd out. When g ( y ) 1, that is g ( y ) is a constant function whose value is always 1, the separable ODE becomes y 0 = f ( x ) , which is sometimes called integrable ODE . The general solution to the integrable ODE is the antiderivative of f ( x ), i.e. y ( x ) = Z f ( x ) dx So if we can ﬁnd the antiderivative of f ( x ) we can ﬁnd the explicit general solution. Example 2.1 . Find solution
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Unformatted text preview: y = x 2 + 3 Solution The solution is, y ( x ) = Z x 2 + 3 dx = 1 3 x 3 + 3 x + C . Here we used the power rule of antiderivative, Z x n dx = x n +1 n + 1 + C. To ﬁnd the solution in Mathcad , you just click any blank area in the workplace and type [Ctrl][I] to get the indeﬁnite operator Z d and type x^2 +3 in the place holder for integrant and x in the inte-grating variable place holder. Then press [Ctrl][.] and click on any region outside the bounding box to get the result. Sometime you might need to hit [F9] too if the auto execution mode is turn oﬀ. a...
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## This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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