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By Belozerova, Tatyana; Henner, Victor; Khenner, Mikhail
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Extra resources for Ordinary and partial differential equations
Sample text
The homogeneous equation is z′ − zx = 0 , or dzz = dxx . Its general solution is z = C1x. The particular solution of the nonhomogeneous equation can be found using the method of the variation of a parameter: z = C1( x )x. Substituting this into equation z′ − z / x = x gives C1′x + C1 − C1x / x = x , C1′ = 1, and C1( x ) = x . Thus z = C1 x + x 2 , and finally y ( x ) = ∫ z( x) dx = ∫ ( x 2 3 2 + C1x ) dx = x + C1 x + C2. 3 2 3. Equation does not contain the independent variabley ( x ). For example, the secondorder equation is F ( y , y ′ , y ′′ ) = 0.
1 presented in the table form. 15) and compares with the reader’s analytical solution (in the * interface it is denoted as x * (t ) and y (t ) , and should be input as shown in the interface). 1) where f ( x , y , y ′ ,…, y (n−1) ) is a given function. 1) depends on n arbitrary constants. 1): y (n ) = f ( x ), a general solution obtained by consequent integration of the equation n times contains n arbitrary constants as the coefficients in the polynomial of order n−1: ∫ ∫ ∫ y( x ) = dx dx … f ( x )dx + n−1 ∑C x .
Indb 29 z( x , y ) = ∫ ϕ(y ) dy +f ( x), 1 14/12/12 11:22 AM 30 ◾ Ordinary and Partial Differential Equations where f1(x) is an arbitrary function of x. Let denote ∫ ϕ(y )dy = f2(y ) —this function is also an arbitrary function of y (since φ (y) is arbitrary). At last, z( x , y ) = f1( x ) + f2(y ). Problems Answers 1. 2 xydx + ( x 2 − y 2 )dy = 0 3 yx 2 − y 3 = C 2. (2 − 9 xy 2 )xdx + (4 y 2 − 6 x 3 ) ydy = 0 x 2 − 3x 3 y 2 + y 4 = C 3. 3x 2 (1 + ln y )dx = (2 y − x 3 / y )dy x 3 (1 + ln y ) − y 2 = C 2 x 2 + ( x 2 − y )3/2 = C 3 4.