Differential Equations

Differential Equations - Contents 15 Differential Equations...

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Unformatted text preview: Contents 15 Differential Equations 222 15.1 Simple Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 15.2 Geometric Aspects of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 15.3 Solving Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 15.4 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 15.5 Linear Equations of First Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 15.6 Homogeneous Linear Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . 230 CONTENTS 2 Chapter 15 Differential Equations 15.1 Simple Differential Equations Verify that the differential equation in question 1 to 16 has the given function as a solution. 1. x dy dx + y = x 2 ; f ( x ) = 1 3 x 2 + c x NOTE: y = f ( x ) and f ( x ) = dy dx Solution: The derivative of f is f ( x ) = 2 x 3- c x 2 . Substitute in for dy dx x 2 3 x- c x 2 + 1 3 x 2 + c x = x 2 . Equation is valid for all x for which f ( x ) is defined, namely x 6 = 0. 2. dy dx = x 2 ; f ( x ) = 1 3 x 3 3. dy dx = 3 x + 5; f ( x ) = 3 2 x 2 + 5 x- 2 4. ds dt = 1 √ t- 1 ; f ( t ) = 2( √ t- 1 + 1) 5. x 3 y = x 4- √ 3; g ( x ) = 1 2 x 2 + √ 3 2 x 2 6. d 2 y dx 2 = x,y = 1 6 ( x 3 + 2 x- 2) 7. d 2 s dt 2- 2 ds dt + 10 = 0; s = 5 t 8. xy = 2 y ; y = Cx 2 9. y 000- y 00 = 0; y = C 1 + C 2 x 10. y dy dx = x 2 , y = q 2 3 x 3 + 4 11. u x du dx = 1; u = √ x 2 + k 12. dy dx = y 2 x 2 ; y = x 1- ax 13. x ( y ) 2- yy + 1 = 0; y = C + x C 223 15.1 Simple Differential Equations 14. dx dt + x t = √ t 2 + 1; x = ( t 2 +1) 3 2 + C 3 t 15. D 2 x y- 3 xD x y + 3 y = 2- 3 x 2 ; y = x 2 + 3 x 16. xD 2 x y- xD x y = 12 x 2- 6 x 3 ; y = 2 x 3 + 5 In problems numbered 17 to 40, find the solution of the differential equation satisfying the initial conditions. 17. y = 2 x- 4; y = 3 at x = 3 Solution: y = R 2 x- 4 dx or y = x 2- 4 x + C where C is an arbitrary constant. To determine C substitute in the initial conditions ( x = 3 , y = 3) 3 = 9- 12 + C or C = 6 . The answer is y = x 2- 4 x + 6. Comment: y = x 2- 4 x + C is the general solution. y = x 2- 4 x + 6 is the particular solution. 18. y = 4 x + 7; y = 3 at x = 2 19. y =- x 2- 3; y = 4 at x =- 1 20. ds dt = 3 t- 5; s =- 3 at t = 2 21. F ( t ) = 1- 6 t + t 2 ; F (- 1) = 4 F ( t ) = t 3- 3 t + t + 25 3 22. dv dt =- 2 t + 3; v = 4 at t = 3 23. dy dx = √ 7 x 3 + 3 x- 2; y =- 5 when x = 0 24. y = ax 2 ; y = a when x = 1 25. 1 t dy dt = 2 t 2 + t + 1; y = 5 when t = 1 2 26. ds dt = √ 1- 2 t ; s = 0 when t =- 4 27. 1 x dy dx = 8 3 √ 1 + 2 x 2 ; y = 2 when x = 0 28. y =- x √ 10- x 2 ; y = 3 when x =- 2 29....
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Differential Equations - Contents 15 Differential Equations...

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