Sec_2.3 - 2.3 Homogeneous Equations If a function f has the...

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Unformatted text preview: 2.3 Homogeneous Equations If a function f has the property that f ( tx, ty ) = t n f ( x, y ) for some real number n , then f is said to be a homogeneous function of degree n . Ex1: Some Homogeneous Functions 2 2 a. f ( x, y ) = x - 3 xy + 5 y The degree of each term is 2, therefore we can say that this function is a homogeneous function of degree 2. 2 2 f ( tx, ty ) = ( tx ) - 3 ( tx ) ( ty ) + 5 ( ty ) f ( tx, ty ) = t 2 f ( x, y ) . b. f ( x, y ) = 3 x 2 + y 2 f ( x, y ) = 3 x 2 + y 2 f ( tx, ty ) = 3 t 2 x 2 + t 2 y 2 f ( tx, ty ) = 3 t 2 ( x 2 + y 2 ) = t 2 / 3 3 x 2 + y 2 f ( tx, ty ) = t 2 / 3 f ( x, y ) . This function is a homogeneous function of degree 2/3. 2 2 c. f ( x, y ) = x + y + 1 f ( x, y ) = x 2 + y 2 + 1 f ( tx, ty ) = t 2 x 2 + t 2 y 2 + 1 Definition 2.2 Homogeneous Function = t 2 x 2 - 3t 2 xy + 5t 2 y 2 = t 2 ( x 2 - 3xy + 5 y 2 ) f ( tx, ty ) t 2 f ( x, y ) . This function is not a homogeneous function. d. f ( x, y ) = x +4 2y f ( x, y ) = x tx x + 4 f ( tx, ty ) = +4= +4 2y 2ty 2y f ( tx, ty ) = t 0 f ( x, y ) . This function is a homogeneous function of degree 0. If a function f is a homogeneous function of degree n , then we can rewrite x y f ( x, y ) = x n f f ( x, y ) = y n f ,1 1, . y x Definition 2.3 Homogeneous Equation A differential equation of the form M ( x, y ) dx + N ( x, y ) dy = 0 is said to be homogeneous if both M and N are both homogeneous functions of the same degree. The method to solve it is to substitute y = ux or x = vy into the differential equation. Note: y and u are functions of x if y = ux is selected; or otherwise. 2 2 2 Ex2. Solve ( x + y ) dx + ( x - xy ) dy = 0 . This is a homogeneous equation. To solve it, the method is to substitute y = ux into the equation, then ( x 2 + y 2 ) dx + ( x 2 - xy ) dy = 0 ( x 2 + u 2 x 2 ) dx + ( x 2 - ux 2 ) ( udx + xdu ) = 0 x 2 ( 1 + u ) dx + x3 ( 1 - u ) du = 0 ( x 2 + u 2 x 2 + ux 2 - u 2 x 2 ) dx + x3 ( 1 - u ) du = 0 dx u - 1 = du x u +1 dx 2 dx 2 = - 1 du = - 1 du x u +1 x u +1 y y ln x + c = u - 2ln u + 1 = - 2ln + 1 x x x y + y ln x + ln + ln c1 = x x ln c1 2 ( x + y) x 2 y = . x ln c1 ( x + y) x 2 2 y ( x + y) = e y / x = c1 x x 2 ( x + y ) = c2 xe y / x . Ex3. Solve 2 xy - y dx - xdy = 0 . Let y = ux , then ( ) (2 xy - y dx - xdy = 0 ( 2u1/ 2 x - ux ) dx - x ( udx + xdu ) = 0 ( 2u1/ 2 - 2u ) xdx - x 2 du = 0 dx du = 1/ 2 . x 2u - 2u It is hard to integrate the equation above, then select 1 du 1 t = u1/ 2 dt = u -1/ 2 du 1/ 2 = dt 2 2u - 2u 1 - t dx du dx 1 = 1/ 2 = dt x 2u - 2u x 1- t ln ( x ) + c = - ln ( 1 - t ) ln ( x ) + ln ( 1 - t ) = c1 x ( 1 - t ) = c2 1 - c 1 - 2 = x y c2 = x x 2 ) y ( x - c2 ) = y. x x 3 4 4 Ex4. Solve 2 x ydx + ( x + y ) dy = 0 . Choose x = vy .Then 2 x3 ydx + ( x 4 + y 4 ) dy = 0 3x 4 y 2 + y 6 = c. Ex. Solve the IVP x Choose y = ux .Then dy = y + xe y / x , y ( 1) = 1 . dx dy = y + xe y / x xdy = ( y + xe y / x ) dx dx x ( udx + xdu ) = ( u + eu ) xdx x x 2 du = eu xdx du dx = eu x du dx = -e - y / x = ln x + c. u e x Substitute the initial condition, then -e- y / x = ln x + c -e -1 = ln ( 1) + c c = -e-1 e -1 - e- y / x = ln x . Homework: 1~43 ODD. ...
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This note was uploaded on 04/01/2008 for the course MATH 2214 taught by Professor Edesturler during the Fall '06 term at Virginia Tech.

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