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Exact Differential Equations The following type of first order differential equations that we’ll be looking at is exact differential equations. What’s exact differential equation? Property: or (1) is an exact differential equation if and only if For exact differential equation (1), there exists a function such that and . Thus, Thus, differential equation (1) becomes So, is an implicit solution of differential equation (1). Let’s look at the detail procedure to find the solution for exact differential equation from the following examples. Example 1: Solve the differential equation: yy2 Let M ycosx2xe ,N sinxx e 1 yy M cosx2xe ,N cosx2xe yx MN yx There exists a function , such that MN, . xy y ycosx 2xe (a) x 2 y sinxx e 1 (b) y Integrating (a) with respect to x, we have yy2 ycosx 2xe dx ysinxx e h(y) Take dirivative with respect to y, we have sinxx2ey h'(y) y From (b), we have sinx x2ey 1, so y 22yy sin x x e h'(y) sin x x e 1, hy'( ) 1 h()y yc Thus, ysin x x2ey yc ysin x x2ey yc 0 is the solution to yy2 ycosx2xe (sinxx e 1)y'0 Example 2: Solve the differential equation: Transfer the original equation to 22 (xy 2)dx(x y3)dy 0 22 Let M xy 2,N x y3 M 2xy,N 2xy yx MN yx There exists a function , such that MN, . xy 2 xy 2 (a) x 2 x y 3 (b) y Integrating (a) with respect to x, we have (xy2 2)dx 1 x2y2 2xh(y) 2 Take dirivative with respect to y, we have 2 y x y h'(y) 2 From (b), we have y xy 3, so 22 x yh'(y) x y3, hy'( ) 3 h(y) 3yc 1 22 Thus, 2 x y 2x3yc 1 22 2 x y 2x3yc0 is the solution to 22 xy 2(3x y)y' Example 3: Solve the differential equation: with initial value: . Transfer the original equation to 2ty ( 2t)dt (ln(t2 1)2)dy 0 t2 1 Let M 2ty 2t,N ln(t2 1)2 t2 1 22tt MN, yt 22 tt11 MN yx There exists a function , such that MN, . xy 2ty 2ta( ) x t2 1 2 y ln(tb1)2 ( ) Integrating (a) with respect to t, we have 2ty 22 ( 2 2t)dt yln(t 1)t h(y) t 1 Take dirivative with respect to y, we have ln(t2 1)h'(y) y From (b), we have ln(t2 1)2, so y 22 ln(t 1)h'(y) ln(t 1)2, hy'( ) 2 h(y) 2yc 22 Thus, yln(t 1)t 2yc 22 yln(t 1)t 2yc0 is the solution to ( 2ty 2t)dt (ln(t2 1)2)dy 0 t2 1 y(1) 0 22 0ln(1 1)1 20cc0 1 22 yln(t 1)t 2y10 is the solution to 22 yln(t 1) t 2yc0 with initial value y(1) 0. Remark: This equation is first order linear, you can use first order linear equation method to solve it too. Integrating Factor: Sometimes it is possible to convert a differential equation that is not exact into an exact equation by multiplying the equation by a suitable integrating factor. ()x Multiply both sides to so that the resulting equation ()x ()x is exact, then we call is an integrating factor. ()x Example 4: Verify that is an integrating factor of the equation ()xx (2) 22 (3xy y )(x xy)y'0 Multiplying to both sides of (2), we have ()xx (3x2y xy2)(x3 x2y)y' 0 Let M 3x2yxy2,N x3x2y, we have 22 M 3x 2xy,N 3x 2xy yx MN yx Differential Equation (2) is exact. Example 5: Given is an integrating factor of the equation ()xx (2) 22 (3xy y )(x xy)y'0 Find the solution to (2). Multiplying to both sides of (2), we have ()xx (3) (3x2y xy2)(x3 x2y)y' 0 (3) is an exact equation, there exists a function , such that 2 2 3 2 M3,x yxy Nx x y xy 3x2yxy2dx x3y1x2y2 h(y) 2 32 y x x yh'(y) 32 y x x y x3x2yh'(y) x3x2y h'(y) c x3y1x2y2c 2 x3y1x2y2 0 is the solution to (3xy y2)(x2 xy)y' 0. 2
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