ME 391 Mechanical Engineering Analysis

Transcription

1 Solve the following differential equations. ME 9 Mechanical Engineering Analysis Eam # Practice Problems Solutions. e u 5sin() with u(=)= We may rearrange this equation to e u 5sin() and note that it is not separable, there is no variable transformation, but since it is linear we can use the Integrating Factor Method. Hence we need to find F and G, so we may transform the equation to the form d F()u G() where e F ep ds ep Ei() where Ei is the Eponential Integral Function. Then sin() sin() G F ep Ei() Then our equation becomes d sin() ep Ei() u ep Ei() which is in separate and integrate form. Separating sin() dep Ei() u ep Ei() Integrating sin() ep Ei() u ep Ei() C Unfortunately, the integral over t does not eist in analytical form, so our final solution is sin() C u ep Ei() ep Ei() ep Ei()

2 ME 9 Mechanical Engineering Analysis u. 7u u with u(=0) = We can rearrange this equation to u u 7u which is clearly separable, so we may use the Separate and Integrate Method. Separating u u 7u Integrating u u 7u lnu 7/ C 6 Probably a good place to apply the initial condition ln() 7/ (0) C 6 ln/ C 6 Now solving for u u 7/ ep 6C 6C 7/ u ep Our final solution becomes u ep ln(/) 7/. ln() 8 with u(=) = 0 We can rearrange this equation to 8 ln() which is clearly separable, so we may use the Separate and Integrate Method. Separating 8 ln() Integrating 8 ln() ln u C

3 ME 9 Mechanical Engineering Analysis Applying the initial condition () ln() u( ) 0 () C Our final solution is ln u C C u. 9u with u(=) = The equation is clearly not separable. However, if we rearrange it by dividing through by u u 9 it should then be obvious that we can use the Variable Transformation Method, where = u/ Applying the transformation 9 We have shown d Substituting we have d 9 Rearranging d 8 which is in separable form. Now separating d 8 And integrating d ln(c) 8 tanh / ln(c) Solving for tanh ln(c) or u tanh ln(c) Applying the condition u( ) () tanh ln(c())

4 ME 9 Mechanical Engineering Analysis Solving for C tanh C ep (.75) / 5. cos()u sin()u / 9ln() with u(=) = This equation is not separable, there does not seem to be an appropriate variable transformation, and since it is nonlinear we cannot use an integrating factor. Let s check to see if it is eact. We have / / N(,u) cos()u and M(,u) sin()u 9ln() Then N / M sin()u and sin()u u So the equation is eact and we can use the Eact Equation Method. We start with f (u) g() N M to find f and g. Substituting for M and N / / f (u) g() cos()u sin()u 9ln() Integrating / / f (u) g() u cos() u cos() ln() f (u) g() ln() So that f (u) 0 g() ln() Substituting into M(,u) f (u) C / u Now rearranging cos() ln() 0 C / u cos() C ln() and solving for u C ln() u cos() Applying the initial condition / C () ln() u( ) cos(()) / C cos() /

5 ME 9 Mechanical Engineering Analysis Solving for C Finally C cos().7.7 ln() u cos() / u u 6. cos(u / ) with u(=) =0 The equation is clearly not separable. However, if we rearrange it by multiplying through by u u cos(u / ) it should then be obvious that we can use the Variable Transformation Method, where = u/. Applying the transformation cos( ) We have shown d Substituting we have d cos( ) Rearranging d cos( ) which is in separable form. Now separating d cos( ) And integrating d ln(c) cos( ) Unfortunately the integral is not analytic, so we must stop at this point. e dt e 7. 7cos() with T(=0) = 6 T T This equation is not separable, there does not seem to be an appropriate variable transformation, and since it is nonlinear we cannot use an integrating factor. Let s check to see if it is eact. We have e e N(,T) and M(,T) 7cos() T T 5

6 ME 9 Mechanical Engineering Analysis Then N e M e and T T T So the equation is eact and we can use the Eact Equation Method. We start with f (T) g() NdT M to find f and g. Substituting for M and N e e f (T) g() dt 7cos() T T Integrating e e f (T) g() 7sin() T T f(t) g() 7sin() So that f (T) 0 g() 7sin() Substituting into M(,T) f (T) C e 7sin() 0 C T Now rearranging and solving for T / e T C 7sin() Applying the initial condition (0) e T( 0) 6 C 7sin(0) Solving for C C 7 Finally 8. e T / 7 7sin() dy cos( ) / We can rearrange this equation to dy 8 e sin() cos() / e sin() 8 with y(=-) = 6

7 ME 9 Mechanical Engineering Analysis which is clearly separable, so we may use the Separate and Integrate Method. Separating 8 e sin() dy cos() Integrating 8 e sin() dy cos() 8 e sin() y C cos() Unfortunately the integral cannot be found analytically, so we have gone as far as we can go. 9. sin()u 5 with u(=) = 0 We may rearrange this equation to sin()u 5 and note that it is not separable, there is no variable transformation, but since it is linear we can use the Integrating Factor Method. Hence we need to find F and G, so we may transform the equation to the form d F()u G() where sin() ((sin()) ) F ep sin() where Ei is the Eponential Integral Function. Then sin() ((sin()) ) G F5 5 Then our equation becomes d sin() ((sin()) ) sin() ((sin()) ) u 5 which is in separate and integrate form. Separating sin() ((sin()) ) sin() ((sin()) ) d u 5 Integrating sin() ((sin()) ) sin() ((sin()) ) u 5 C sin() ((sin()) ) 5sin() 5 cos() 5 sin() u C 6 8 7

8 ME 9 Mechanical Engineering Analysis Solving for u 5sin() 0 cos() 0 sin() C u sin() ((sin()) ) sin() ((sin()) ) Applying the condition 5sin() 0 cos() 0 sin() C ( ) 0 sin() ((sin( )) ) sin() ((sin( )) C 0 0 ((0) ) 0 ((0) ) 5 C 0 Solving for C 5 C 8 Then our final solution becomes 5sin() 0 cos() 0 sin() 0 u sin() ((sin()) ) 0. ) u cos( u) 7t t with u(t=) = - Rearranging the equation cos(u) t 7t which is separable, so we can use the Separate and Integrate Method. Separating cos(u) t 7t Integrating cos(u) t 7t sin( u) 7 t t C Solving for u 9 u sin t t C Applying the initial condition 9 u(t ) sin () () Solving for C sin( 9) C Our final solution is 9 u sin t t sin( 9) 6 8 C

9 ME 9 Mechanical Engineering Analysis d u. 6u 0 with u(=) = and 0 0 We follow our step by step method. Step Equation is in the proper form. Step Substituting e m, m m 6 0 Step Solve for the roots of m. m m So that a - b < 0 Step Not Applicable Our solution becomes 0.5 u() e Acos Not Applicable 5.75 Bsin 5.75 Step 7 Applying our conditions to solve for and B 0.5 u( ) e Acos 5.75 Bsin 0.5 e A 5.75 sin 5.75 B cos e Acos 5.75 Bsin 5.75 at =0 0 B A Solving for A A 5.75B Substituting into our first condition e Bcos Bsin

10 ME 9 Mechanical Engineering Analysis Solving for B e B 5.75 cos So that our final solution is e u() 0.5() sin cos 5.75 cos 5.75 sin sin 5.75 d u. u 0 with u(=0) = and u(=) = 0 We follow our step by step method. Step Equation is in the proper form. Step Substituting e m, m m 0 Step Solve for the roots of m. m 6 6 m 6 6 So that a - b = 0 Step Not Applicable Not Applicable Our solution becomes u() (C C ) e Step 7 Applying our conditions to solve for C and C u( 0) C u( ) 0 ( C C = - So that our final solution is u() ( )e ) e 0

11 ME 9 Mechanical Engineering Analysis d u, 8 5u 0 with u(=0) = 0 and 0 We follow our step by step method. Step Equation is in the proper form. Step Substituting e m, m m.5 0 Step Solve for the roots of m. m m So that a - b > 0 Step () C ep C ep.5 u Not Applicable Not Applicable Step 7 Applying our conditions to solve for C and C u( 0) 0 C C 0.775C ep C ep C.5C 0 Substituting for C 0.775C.5C C = 0.08 C = So that our final solution is u() 0.08ep ep.5. Determine the position of the mass as a function of time for a spring-mass system with the following properties:

12 ME 9 Mechanical Engineering Analysis mass, M =.5 kg viscous damping, C =. kg/s spring molus, k = N/m forcing function, F(t) = 0 N dy initial conditions: y(t=0) = 0. m and 0 t0 We begin by writing our general for a mass-spring system ODE d y dy M C ky F(t) Setting F(t) = 0 and rearranging d y C dy k y 0 M M We recognize this equation as being in a form that we can apply the e m method where a = C/M =./.5 =.8 and b = k/m = /.5 = Then we can follow our step by step procere. Step Satisfied Step m + am + b = 0 Step.8 m (.8) () m (.8) (). 0.0 So that a b < 0 Step y(t) e.t Acos 0.0t Bsin0.0t Step 7 At t = 0 y(t 0) 0. A

13 ME 9 Mechanical Engineering Analysis dy e dy t0 B =.5.t t 0.0Asin0.0t 0.0Bcos0.0t.e. Acos 0.0t Bsin0.0t 0.0B.A So our final solution is.t y(t) e 0.cos 0.0t.5sin0.0t 5. Determine the position of the mass as a function of time for a spring-mass system with the following properties: mass, M = 0.75 kg viscous damping, C =. kg/s spring molus, k = N/m forcing function, F(t) = 0 N dy initial conditions: y(t=0) = 0 m and 0. m/s t0 We begin by writing our general for a mass-spring system ODE d y dy M C ky F(t) Setting F(t) = 0 and rearranging d y C dy k y 0 M M We recognize this equation as being in a form that we can apply the e m method where a = C/M =./0.75= 5.6 and b = k/m = /0.75 = Then we can follow our step by step procere. Step Satisfied Step m + am + b = 0 Step 5.6 m (5.6) () m (5.6) ().76 So that a b > 0 Step u() C e 0.8t.76t Ce

14 ME 9 Mechanical Engineering Analysis Step 7 At t = 0 y(t 0) 0 C C dy 0.8t 0.8Ce.76Ce dy C.76C t0 C = -C C.76C C = C = 0.05 So our final solution is 0.8t.76t u() 0.05e 0.05e.76t d u 6. 5u 5 with u(=0) = 0 and 0 0 This is a nd order, liner, constant coefficient, nonhomogeneous ODE. We know that the solution is composed of a homogenous part and a particular part or u() = u h + u p For the homogenous solution we will use the e m method and for the particular solution we use the method of undetermined coefficients. Beginning with the homogenous solution. Step Satisfied. Step m m 5 0 Step Solve for the roots of m. For a nd order equation, we will have m () (5) m () (5) Hence a b < 0 Step

15 ME 9 Mechanical Engineering Analysis Our homogenous solution becomes () e C cos C sin uh Step 7 Skip until we combine with the particular solution Now we apply the method of undetermined coefficients Step Since our forcing function is + 5, we assume a polynomial function for our particular solution of the form u p () = A + B + C Step Substituting into our differential equation A A B 5 A B C 5 Step Collecting like powers of terms : 5A = : 8A + 5B = 0 0 : A + B + 5C = 5 Step Solving for the undetermined coefficients A=0. B = -8A/5 = -0. C = (5-A -B)/5 =.76 Assemble out total solution u() e C cos C sin

16 ME 9 Mechanical Engineering Analysis Apply our conditions u( 0) 0 C.76 C.76 e C sin C cos e C cos C sin C C 0. 0 C = C + 0. = (-.76) + 0. = -.0 Then our final solution becomes u() e.76cos.0sin d u 7. 7 u 5cos() with u(=0) = 0 and u(=) = This is a nd order, liner, constant coefficient, nonhomogeneous ODE. We know that the solution is composed of a homogenous part and a particular part or u() = u h + u p For the homogenous solution we will use the e m method and for the particular solution we use the method of undetermined coefficients. Beginning with the homogenous solution. Step Satisfied. Step m 7m 0 Step Solve for the roots of m. For a nd order equation, we will have 7 m (7) () m (7) () 6.70 Hence a b > 0 Step Our homogenous solution becomes u () C e C e h 6

17 ME 9 Mechanical Engineering Analysis Step 7 Skip until we combine with the particular solution Now we apply the method of undetermined coefficients Step Since our forcing function is + 5cos(), we assume the following function for our particular solution u p () = A + B + Ccos() + Dsin() Step Substituting into our differential equation 6Ccos() 6Dsin() 7A Csin() Dcos() A B Ccos() Dsin() 5cos() Step Collecting like terms of : A = 0 : 7A + B = 0 cos(): -6C+8D+C = 5 sin(): -6D 8C + D = 0 Step Solving for the undetermined coefficients A = B = -.5 C = D = 0. Assemble out total solution 0.98 u() C e C e cos() + 0.sin() Apply our conditions u( 0) 0 C C u( ) Ce Ce cos() + 0.sin() from which we can solve for C and C. 7

18 ME 9 Mechanical Engineering Analysis 8. Determine the position of the mass as a function of time for a spring-mass system with the following properties: mass, M = 0.75 kg viscous damping, C =. kg/s spring molus, k = N/m forcing function, F(t) = 5e.t N dy initial conditions: y(t=0) =0. m and 0. m/s t0 We begin by writing our general for a mass-spring system ODE d y dy M C ky F(t) Setting F(t) and rearranging d y C dy k 5.t y e M M M where a = C/M =./0.75= 5.6 and b = k/m = /0.75 = and 5/M = 5/0.75 = This is a nd order, liner, constant coefficient, nonhomogeneous ODE. We know that the solution is composed of a homogenous part and a particular part or y(t) = y h + y p For the homogenous solution we will use the e m method and for the particular solution we use the method of undetermined coefficients. Beginning with the homogenous solution. Then we can follow our step by step procere. Step Satisfied Step m + am + b = 0 Step 5.6 m (5.6) () m (5.6) ().76 So that a b > 0 Step y (t) C e h 0.8t C e.76t 8

19 ME 9 Mechanical Engineering Analysis Step 7 Skip until we add the particular solution Now we apply the method of undetermined coefficients Step Since our forcing function is 6.67e.t, we assume the following function for our particular solution y p (t) = Ae.t Step Substituting into our differential equation.t.t.t (.) Ae 5.6(.)Ae Ae 6.667e Step Collecting like terms of.5a + 6.7A + A= Step Solving for the undetermined coefficients A = Assemble out total solution 0.8t y(t) C e C e.76t.t 0.578e Apply our conditions y(t 0) 0. C C dy 0.8t.76t 0.8C e.76ce 0.68 dy C.76C 0.68 t0 from which we can solve for C and C..t e.t 9