0.1 Problems to solve

Transcription

1 0.1 Problems to solve Homework Set No. NEEP 547 Due September 0, 013 DLH Nonlinear Eqs. reducible to first order: 1. 5pts) Find the general solution to the differential equation: y = [ 1 + y ) ] 3/. 5pts) page 7, prob. 13c; Find the general solution to the differential equation: y y = y y + y ) Linear Operators 3. 6pts) First factor the equation using operator notation and then find the general solution to the differential equation: x d y dx + x dx y = pts) First factor the equation using operator notation and then find the general solution to the differential equation: x d y dx + dx = 3 x x Linear dependent or independent solutions. 5. 4pts) page 69, prob. 8. : Show that y 1 x) = x and y x) = x are linearly independent solutions of x y x y + y = 0 on [-1,1], but that W 0) = 0. Why does this not contradict Theorem.3.1 in this interval? Theorem.3: Wronskian Test : Let y 1 and y be solutions of y + px) y + qx) y = 0 on the open interval I. Then,.3.1. Either W x) = 0 for all x in I, or W x) 0 for all x in I..3.. y 1 and y are linearly independent on I if and only if W x) 0 on I. 6. 4pts) page 69, prob. 10: Show that y 1 x) = 3 e x 1 and y x) = e x + are solutions of y y + y y ) = 0, but neither y 1 nor y 1 +y is a solution. Why does this not contradict Theorem.? Theorem.: Let y 1 and y be solutions of y + px) y + qx) y = 0 on an interval I. Then any linear combination of these solutions is also a solution. Homogeneous Linear Differential Equations with Constant Coefficients: 7. 6pts) Solve the initial-value problem: D 3 6 D + 11 D 6) y = 0 where D n = dn dx n ; with conditions: y = y = 0 and y = when x = pts) Solve the initial-value problem: 8y 4y + 6y + 5y = 0 with conditions: y = 0, y = y = 1 when x = 0. Nonhomogeneous Equations with Constant Coefficients 9. 6pts) O Neil, page 93 prob. 16; find the general solution: y y + y = 3 x + 5 sin3x) 10. 7pts) find the general solution: y iv + 3 y 4 y = sinhx) sin x) 1

2 0. Problem 1 reduction of order book.3 section) Nonlinear Eq, reducible to first order. Find the general solution to y = 1 + y ) ) 3 This is non-linear, second order differential equation. Since x does not appear explicitly, let u = y, then u = y and the above differential equation becomes The above is now separable and solved for u The above is solved explicitly for u Since u = y therefore u u = 1 + u ) 3 du = dx 1 + u ) 3 u = x + C 1 + u = x + C) 1 + u u = 1 + u ) x + C) u u x + C) = x + C) This is separable, hence the solution is y x) = ± u x + C) = 1 x + C) x + C) u = ± 1 x + C) = ± x + C) 1 x + C) y x + C) = ± 1 x + C) x + C) 1 x + C) + C = ± 1 x + C) + C = ± 1 x xc + C + C 0.3 Problem O Neil page 7, problem 13c Find general solution to yy = y y + y ) Solution: This is non-linear, second order differential equation. ) y d y = y dx dx + dx d y dx = y dx + 1 ) y dx

3 Multiply by dx d y dx dx = y + 1 y dx Let u y) = dx u here is function of y.the differential equation becomes Multiply by dx Hence and using that u = dx du 1 dx u = y + 1 y u du dx = yu + 1 y u the above reduces to du dx dx = yudx + u y du = y + 1 y = y + 1 y = y + 1 y u du u y = y dx ) dx dx ) dx This is solved for u y).the integrating factor is I f = y 1 hence But u = dx hence d y 1 u ) = y 1 y = 1 y 1 u = y + C 1 u = y + C 1 y dx = y + C 1 y dx y C 1 y = 0 This is first order non-linear ODE. It is separable, hence Applying partial fractions to the LHS gives y + C 1 y) = dx C 1 y 1 C 1 C 1 + y = dx 3

4 Integrating both sides now gives 1 C 1 ln y 1 C 1 ln y + C 1 ) = x + C ln y ln y + C 1 ) = C 1 x + C 3 Where C 3 = C 1 C ln y y + C 1 = C 1 x + C 3 y y + C 1 = C 4 e C 1x y = yc 4 e C 1x + C 1 C 4 e C 1x y yc 4 e C 1x = C 1 C 4 e C 1x y 1 C 4 e C 1x ) = C 1 C 4 e C 1x y = C 1C 4 e C 1x 1 C 4 e C 1x 0.4 problem 3, linear operators First factor the equation using operator notation and then find the general solution x y + xy y = 0 Let D d dx.the ODE can be written as x D + xd 1 ) y = 0 The roots of the characteristic equation x λ + xλ 1 are m = b± b 4ac 1± 5 x = 1 ± 5 x x The ODE becomes Let hence. Hence the roots are m 1 = x m = 1 5 x D m 1 ) D m ) y = 0 = x± x +4x = x±x 5 = a x x D m ) y = v 1) D m 1 ) v = 0 Solution of D m) v = 0 is solution of v mv = 0 which is v x) = Ae mx hence the solution of the above becomes v x) = Ae m 1x = Ae = Ae ) 1+ 5 x x )

5 Hence v x) is constant and does not depend on x. Let Ae ) 1+ 5 D m ) y = v = C 1 dx m y = C 1 dx x y = C 1 The solution to the homogenous equation is h dx x y = 0 h y = x dx ) 1 5 ln y h = y h = C xe y h = C 3 x 1 ) 5 = C 1.Now from Eq. 1) ln x + C For the particular solution, using the trial y p = C, hence p m dx y p = C 1 or C = C x 1, hence C 1 = C 4 x, so Therefore the general solution is y p = C 4 x y = y h + y p = C 3 x + C 4 x Where C 4, C 3 are constants that can be determined from initial or boundary conditions 0.5 problem 4 First the homogenous equation is solved. xy + y = 3x x Let D d dx hence xd + D ) y h = 0 The roots of the characteristic equation xλ + λ) are m = b± b 4ac are Therefore m 1 = x m = 1 1 x = 0 = 1 x = 1± 1 = 1±1 a x x hence the roots D m ) D m 1 ) y h = 0 D m ) D) y h = 0 1) 5

6 Let Hence D) y h = v ) D m ) v x) = 0 dv dx m v x) = 0 dv v = m dx 1 ln v = x dx + C ln v = ln x + C Hence Where C 1 is new constant. Eq. ) becomes v x) = C 1 x To find particular solution, let D) y h = v = C 1 x y h = C 1 x h = C 1 x dx y h = C 1 ln x + C y p = ax 3 + bx + cx + d and y p = 3ax + bx + c and y p = 6ax + b hence the original ODE becomes x 6ax + b) + 3ax + bx + c ) = 3x x Hence c = 0 and a = 1 3 and 4b = 1 or b = 1 4, therefore And the full solution is y = y h + y p 9ax + 4bx + c = 3x x y p = 1 3 x 1 4 x = C 1 ln x + C x 1 4 x 0.6 Problem 5 linear dependent and independent solution) Problem page 69, problem 8 Show that y 1 x) = x and y x) = x are linearly independent solutions to x y xy + y = 0 on [ 1, 1] but that W 0) = 0. Why does this not contradict theorem.3.1 in this interval? Theorem.3: Wronskian test: Let y 1, y be solutions of y + p x) y + q x) y = 0 on the open interval I, then the following is true 6

7 1. Either W x) = 0 for all x in I, or W x) 0 for all x in I. y 1 and y are linearly independent on I iff W x) 0 on I Answer: First we show that y 1, y are solutions to the ODE. Looking at y 1, then y 1 = 1, y = 0. Substituting into the ODE gives x + x = 0 Hence y 1 is a solution. Looking now at y, then y = x, y =. Substituting into the ODE gives x 4x + x = 0 Hence y is also a solution. Now we will show they are linearly independent. Let ay 1 + by = 0 Where a, b are constants. If there are non-zero constants a, b that will make the above true, then y 1, y are linearly dependent. Another way to say this, is that if and only if when a = b = 0 then the above is true, then y 1, y are linearly independent. Let us assume that for all x the following is true ax + bx = 0 Let x = 1, then a + b = 0. Let x = 1 then b a = 0. Solving for a, b from these two equations shows that b = 0 or b = 0, hence a = 0. Therefore, for ay 1 + by to be zero then a = b = 0. This shows that y 1, y are linearly independent. The above showed that y 1, y are solutions to the ODE and that they are linearly independent functions. Now the Wronskian test is applied y 1 y W = y 1 y = x x 1 x = x x = x At point 0 we see that W 0) = 0. This seems like a conflict. But the Abel s stronger statement applies only for solutions of an ODE, which says that for second order ODE, if y 1, y are linearly independent solutions of the ODE, then W can not be zero at any point in the interval. However, there is no conflict in this case, since at x = 0 this statement does not even apply, as we see that when x = 0 the first and second terms of the ODE itself vanish and we no longer have an ODE in first place. At any other point x, where the ODE remain in effect as stated, then W x) 0, and hence there is no conflict. Summary: To show that two functions are linearly independent on an interval, it is enough to show that the W is not zero on any one point in the interval. We do not need to check at each point. It is only when these two functions are also solutions of the ODE, then we need to check that W is not zero on each point in the interval, where the ODE is defined. In this problem, it happened that at x = 0 the ODE itself is not defined since a 0 = 0 there. 0.7 Problem 6 page 69 problem 10 Show that y 1 x) = 3e x 1 and y x) = e x + are solutions of yy + y y ) = 0 but neither y 1 nor y 1 + y is a solution. Why does this not contradict theorem.? Theorem.: Let y 1, y be solutions of y + p x) y + q x) y = 0 on interval I, then any linear combination of these solutions is also a solution. 7

8 Solution First we show that the y 1 and y are solutions. This is done by substitution into the ODE and checking for identity. Starting with y 1 y 1 = 6e x, y 1 = 1e x, hence the ODE become y 1 y 1 + y 1 y 1) = 3e x 1 ) 1e x) + 6e x) 6e x) = 36e 4x 1e x + 1e x 36e 4x = 0 This shows that y 1 is a solution. Now for y we have y = e x, y = e x, hence the ODE become y 1 y 1 + y 1 y 1) = e x + ) e x) + e x) e x) = e x + e x e x e x = 0 Therefore y is also a solution. now to Check if y 1 is a solution. Let y 3 = y 1 = 6e x hence y 3 = 1e x and y 3 = 4e x. Substitution into the ODE gives y 3 y 3 + y 3 y 3) = 6e x ) 4e x) + 1e x) 1e x) = 144e 4x 48e x + 4e x 144e 4x = 4e x 0 Hence y 3 = y 1 is not a solution. Now to check that y 1 + y is a solution or not. Let y 4 = y 1 + y = 3e x 1 + e x + = 3e x + e x + 1, hence y 4 = 6e x e x and y 4 = 1e x + e x, and substitution into the ODE gives y 4 y 4 + y 4 y 4) = 3e x + e x + 1 ) 1e x + e x) + 6e x e x) 6e x e x) = 36e 4x + 3e x + 1e x + e x + 1e x + e x + 1e x e x 36e 4x e x + 1e x = 7e x e x + 4e x 0 Hence y 4 = y 1 + y is not a solution. Now to answer the question. Since the ODE given is not linear, and not of the form y + p x) y + q x) y = 0, then we need to check first that when using the solution y 1 or y 1 + y, the ODE remains of the same form shown above for these to be also solutions. Let us try y 1 and substituting this into the ODE. This results in Dividing by 4 yy + y y ) = 0 y 1 ) y 1 ) + y 1 ) [ y 1 ) ] = 0 y 1 ) y 1 + y 1) y 1) = 0 4y 1 y 1 + 4y 1 4 y 1) = 0 y 1 y 1 + y 1 y 1) = 0 Comparing this with the original ODE, we see it is not the same ODE. The second term was y and now it is y 1. Hence y 1 is not a solution. The reason is due to the nonlinearity of the ODE, the theorem did not apply to it. 8

9 Checking now for the second trial solution y 1 + y and substituting this into the ODE yy + y y ) = 0 y 1 + y ) y 1 + y ) + y 1 + y ) [ y 1 + y ) ] = 0 y 1 + y ) y 1 + y ) + y 1 + y ) y 1 + y ) = 0 y 1 + y ) y 1 + y ) + y 1 + y ) y 1) y ) y 1y = 0 y 1 y 1 + y 1 y ) + y y 1 + y y ) + y 1 + y y 1) y ) y 1y = 0 [ y 1 y 1 + y 1 y 1) ] [ + y y + y y ) ] + y 1 y + y y 1 y 1y = 0 The terms in square brackets are zero, since they are solutions of the ODE and hence vanish, hence the above reduces to y 1 y + y y 1 y 1y = 0 This is not the same ODE we started with. For y 3 = y 1 + y to be a solution, the ODE obtain y 3 y 3 + y 3 y 3) = 0. The reason is due to the nonlinearity of the ODE. 0.8 Problem 7 Solve the IC problem D 3 6D + 11D 6) y = 0 with IC y = y = 0 and y = when x = 0 We need to factor the characteristic equation λ 3 6λ + 11λ 6 = 0. Guessing a root, we see that λ = is a root. Long division gives λ3 6λ +11λ 6 = λ 4λ + 3, hence the characteristic equation is λ λ 4λ + 3) λ ). Now we factor the quadratic giving the final answer of λ 1) λ 3) λ ). The ODE is now written as D 1) D 3) D ) y = 0 Let D ) y = v then Let D 3) v = u then Backtracking to the previous ODE D 1) D 3) v = 0 D 1) u = 0 u u = 0 du dx = u x) ln u = x + c 1 u e x D 3) v = u dv dx 3v e x 9

10 Integrating factor is I f = e 3x hence Now backtracking to the first ODE d dx I fv) = I f c 1 e x I f v = I f c 1 e x dx + c e x dx + c ) 1 e x + c v = c 1 e 3x e x + c e 3x = c 1 ex + c e 3x D ) y = v dx y = c 1 ex + c e 3x Integrating factor is I f = e x hence ) d dx I c1 fy) = I f ex + c e 3x ) e x y = e x c1 ex + c e 3x dx + c 3 ) c1 = e x + c e x dx + c 3 or letting c 1 new constant) then e x + c e x + c 3 y ex + c e 3x + c 3 e x y x) e x + c e 3x + c 3 e x Now the constants are found from IC. y = y = 0 and y = When x = 0 then y = 0, hence 0 + c + c 3 1) Taking derivative, then Hence Taking derivative again At x = 0 y x) e x + 3c e 3x + c 3 e x 0 + 3c + c 3 ) y x) e x + 9c e 3x + 4c 3 e x + 9c + 4c 3 3) Solving Eqs. 1),),3) for the constants gives c 1 = 1, c = 1, c 3 =. The final solution is y x) = e x + e 3x e x 10

11 0.9 Problem 8 Solve the IC problem 8y 4y + 6y + 5y = 0 with IC y = 0, y = y = 1 when x = 0 Solution: Writing the ODE as 8D 3 4D + 6D + 5) y = 0. The first step is to factor the characteristic equation 8λ 3 4λ + 6λ + 5 = 0. By guessing an initial root as λ = 1 with some trials, now performing long Division to reduce it to a quadratic and then applying the quadratic equation to obtain the remaining two roots. Hence 8λ3 4λ +6λ+5 = 8λ 8λ λ+ 1 The characteristic equation now becomes λ + ) 1 8λ 8λ + 10). Factoring the quadratic gives b± b 4ac = 8± 64 48)10) = 8± = 8±16i = 1±i a becomes D 1 + i )) D Let D + ) 1 y = v The ODE becomes )) )) 1 1 D + i D i v = 0 Let D 1 i)) v = u. The ODE becomes 1 D du dx. This means the roots are 1 ± i. Hence the ODE )) 1 i D + 1 ) y = 0 )) + i u = 0 ) 1 + i u = 0 This is separable with solution u e 1 +i)x backtracking to the previous ODE and solving )) 1 D i v e 1 +i) x ) dv 1 dx i v e 1 +i) x Integrating factor is I f = e 1 i)x hence d dx I fv) = I f c 1 e 1 +i) x ve 1 i)x e 1 i)x e 1 +i)x dx + c e [ 1 i) + 1 +i)]x dx + c e ix dx + c Therefore e ix + c v x) e ix e 1 i)x + c e 1 i) x e i+ 1 ) x + c e i+ 1 ) x 11

12 Where c 1. Backtracking to the first ODE, we now solve D + 1 ) y = v dx + 1 y e ) 1 i+ x + c e i+ 1 ) x The integrating factor is e 1 x hence d dx I fy) = I f c 1 e i+ 1 ) x + c e i+ 1 ) ) x I f y = e 1 c x 1 e i+ 1 ) x + c e i+ 1 ) ) x dx + c 3 e i+1)x + c e i+1)x dx + c 3 Therefore e 1+i)x 1 + i + c e 1 i)x 1 i + c 3 e 1+i)x y 1 + i e 1 x e 1 i)x + c 1 i e 1 x + c 3 e 1 x ) = e 1 x e ix c i + c e ix + c 3 e 1 x 1 i But e ix = cos x + i sin x and e ix = cos x i sin x, hence combining the above gives ) y = e 1 x cos x + i sin x cos x i sin x c 1 + c + c 3 e 1 x 1 + i 1 i = e 1 x c1 1 i) cos x + i sin x) + c 1 + i) cos x i sin x) 1 + i) 1 i) ) + c 3 e 1 x ) = e 1 x c1 cos x + i sin x i cos x + i sin x)) + c cos x i sin x + i cos x i sin x)) ) = e 1 x c1 cos x + i sin x i cos x + sin x) + c cos x i sin x + i cos x + sin x) ) = e 1 cos x x c1 ic 1 + c + ic ) + sin x c 1 + ic 1 ic + c ) + c 3 e 1 x Let c 1 ic 1 +c +ic ) = c 4 and let c 1+ic 1 ic +c ) = c 5, then the above reduces to y = e 1 x c 4 cos x + c 5 sin x) + c 3 e 1 x + c 3 e 1 x + c 3 e 1 x This is the general solution. c 3, c 4, c 5 are found from IC. y = 0, y = y = 1 When x = 0 and y = 0 0 = c 4 + c 3 1) Now y = 1 e 1 x c 4 cos x + c 5 sin x) + e 1 x c 4 sin x + c 5 cos x) 1 c 3e 1 x Hence at x = 0 1 = 1 c 4 + c 5 1 c 3 ) 1

13 and y = 1 4 e 1 x c 4 cos x + c 5 sin x) + 1 e 1 x c 4 sin x + c 5 cos x) + 1 e 1 x c 4 sin x + c 5 cos x) + 1 e 1 x c 4 cos x c 5 sin x) c 3e 1 x Hence at x = 0 1 = 1 4 c c c 5 1 c c 3 = 1 4 c c 4 + c 5 3) Solving Eqs. 1),),3) for the constants gives c 3 = 0, c 4 = 0, c 5 = 1, hence the solution is y = e 1 x sin x A plot of the solution is Figure 1: plot of solution to problem 8 HW 0.10 Problem 9, Nonhomogeneous equations with constant coefficients O Neil. page 93, problem 16. Find general solution to y y + y = 3x + 5 sin 3x) Write as D D + 1) y = 3x + 5 sin 3x), where L D D + 1 = D 1) D 1). This will be solved two ways. The first using variation of parameters to obtain the particular solution, and the second by finding particular solution to each separate forcing function and adding First method variation of parameters) Let D 1) y h = v, then D 1) D 1) y h = 0 D 1) v = 0 dv dx v = 0 13

14 Solution is v e x. We now backtrack and solve D 1) y h = v 1,h dx y h e x Integrating factor is e x hence d dx I fy h ) = e x c 1 e x ) e x y h x + c y h xe x + c e x Hence y 1 = xe x and y = e x are the two linearly independent solutions of the homogenous ODE. Let the particular solution be y p = u 1 y 1 + u y where u 1 x), u x) are functions of x to be found. Hence and y p = u 1y 1 + u 1 y 1 + u y + u y y p = u 1y 1 + u 1y 1 + u 1y 1 + u 1 y 1 + u y + u y + u y + u y Therefore, the ODE y p y p + y p = 3x + 5 sin 3x) becomes Collecting terms u 1y 1 + u 1y 1 + u 1y 1 + u 1 y 1 + u y + u y + u y + u y u 1y 1 + u 1 y 1 + u y + u y ) + u 1 y 1 + u y = 3x + 5 sin 3x) u 1 [y 1 y 1 + y 1 ]+u [y y + y ]+u 1y 1 +u 1y 1+u 1y 1+u y +u y +u y u 1y 1 + u y ) = 3x+5 sin 3x) But terms in brackets vanish since this is the ODE with the homogeneous solutions, hence the above reduces to If u then the above becomes 1y 1 + u 1y 1 + u 1y 1 + u y + u y + u y u 1y 1 + u y ) = 3x + 5 sin 3x) {}}{ u 1y 1 + u 1y 1 + u y + u y +u 1y 1 + u y u 1y 1 + u y ) = 3x + 5 sin 3x) d dx u 1y 1 + u y ) + u 1y 1 + u y ) u 1y 1 + u y ) = 3x + 5 sin 3x) u 1y 1 + u y = 0 1) u 1y 1 + u y ) = f x) = 3x + 5 sin 3x) ) Hence we have two equations Eqs. 1),) to solve for u 1, u y u 1 = f x) dx = y 1 y y y 1 y f x) dx W x) 14

15 But Hence and y 1 y W x) = y 1 y = xe x e x + xe x e x e x e x u 1 = 3x + 5 sin 3x)) dx ex = e x 3x + 5 sin 3x)) dx = 3 xe x + 5 e x sin 3x) dx = xex e x + xe x) = e x = e x 3 3x 15 cos 3x) 5 sin 3x) y 1 u = f x) dx W x) xe x = 3x + 5 sin 3x)) dx ex = xe x 3x + 5 sin 3x)) dx = 3 x e x dx 5 e x x sin 3x) dx = e x 6 + 6x + 3x cos 3x + x cos 3x sin 3x + 5 x sin 3x ) ) Therefore y p = u 1 y 1 + u y = e x [ 3 3x 15 + ] sin 3x) xe x cos 3x) 5 e x 6 + 6x + 3x cos 3x + x cos 3x sin 3x + 5 x sin 3x = 3x 3x 15 x cos 3x) 5 x sin 3x) x + 3x cos 3x + x cos 3x sin 3x + 5 x sin 3x = 3x + 3 cos 3x sin 3x + 6 Hence the total solution is )) e x y = y h + y p xe x + c e x + 3x + 3 cos 3x sin 3x

16 0.10. Second method using linearity to add the two separate particular solutions) This will be solved by breaking the forcing functions and solving for each separately and then adding the solutions at the end since the ODE is linear. Hence we will solve the following two ODE s y 1 y 1 + y 1 = 3x y y + y = 5 sin 3x) and the solution will be y = y 1 + y. Starting with the first one, we solve for the homogeneous and then for the particular. D 1) D 1) y 1,h = 0 Now we processed as before. Let D 1) y 1,h = v, then Solution is v e x. We now backtrack and solve D 1) v = 0 dv dx v = 0 D 1) y 1,h = v Integrating factor is e x hence 1,h dx y 1,h e x d dx I fy 1,h ) = e x c 1 e x ) e x y 1,h x + c y 1,h xe x + c e x Now we find the particular solution y 1,p. Let y 1,p = ax + bx + c, hence y 1,p = ax + b and y 1,p = a, therefore the ODE becomes a ax + b) + ax + bx + c = 3x x a) + x 4a + b) + a b + c = 3x Hence a = 0 and b + c = 0 and b = 3. Therefore c = 6 and the forcing function is y 1,p = 3x + 6, hence We now solve the second ode y 1 xe x + c e x + 3x + 6 y y + y = 5 sin 3x) The homogenous solution is the same as above, y,h xe x + c e x. Only the particular solution needs to be found again. Let y,p = A sin 3x + B cos 3x, hence y,p = 3A cos 3x 3B sin 3x and y,p = 9A sin 3x 9B cos x. The ODE becomes 9A sin 3x 9B cos x 3A cos 3x 3B sin 3x) + A sin 3x + B cos 3x = 5 sin 3x) sin 3x 9A + 6B + A) + cos 3x 9B 6A + B) = 5 sin 3x) Therefore, 8A + 6B) = 5 and 8B 6A) = 0, from the first equation A = 6B 5, and from the 8 second 8B 6 6B 5 = 0 or 64B 36B = 0 or B = 1.5, hence A = 9 5 =, therefore 8 8 y,p = sin 3x + 3 cos 3x 16

17 And the general solution is y xe x + c e x + 3x + 6 sin 3x + 3 cos 3x This answer matches the answer obtained above using variation of parameters Problem 10 Find general solution y 4) + 3y 4y = sinh x) sin x First the homogenous solution is find using the operator method. Let D 4 + 3D 4 ) y = sinh x) sin x The characteristic equation is λ 4 + 3λ 4 = 0. Let λ = u, hence u u 4 = 0, and the roots are u = {1, 4}. Hence when u = 1, λ = ±1 and when u = 4, λ = ±i, therefore we obtain the 4 roots as {1, 1, i, i} and the factorization is Solving the homogenous part first. D 1) D + 1) D i) D + i) y = sinh x) sin x D 1) D + 1) D i) D + i) y = 0 Let D + i) y = v, hence Let D i) v = u hence Let D + 1) u = r hence D 1) D + 1) D i) v = 0 D 1) D + 1) u = 0 D 1) r = 0 dr dx r = 0 And the solution is r x) e x, backtracking now we solve Integration factor is e x, hence D + 1) u e x du dx + u e x d dx ex u) = e x c 1 e x ) e x u e x dx + c Therefore e x + c e x u + c e x 17

18 Let c 1 = 1 c 1, hence Backtracking, we now solve u e x + c e x D i) v = u dv dx iv e x + c e x Integration factor is e ix hence d e ix v ) = e ix c 1 e x + c e x) dx e ix v = e ix c 1 e x + c e x) dx + c 3 e ix+x dx + c e ix x dx + c 3 Hence e ix x e ix+x i c i 1 + c 3 Now we backtrack one last time and solve for y h v = e ix e ix+x c 1 i e ix x eix c i 1 + eix c 3 e x i c e x i 1 + eix c 3 D + i) y h = v h dx + iy e x h i c e x i 1 + eix c 3 Integration factor is e ix hence ) d ) e ix y h = e ix e c x 1 dx i c e x i 1 + eix c 3 ) e ix y h = e ix e x c 1 i c e x i 1 + eix c 3 dx + c 4 c 1 = e x+ix c dx + e x+ix dx + e 4ix c 3 dx + c 4 i + 1 i 1 Hence c 1 e x+ix e x+ix = i i c i i + c 3 4i e4ix + c 4 5 ex+ix c 5 e x+ix + c 3 4i e4ix + c 4 y h = e ix c1 5 ex+ix c 5 e x+ix + c 3 4i e4ix + c 4 ) 5 ex c 5 e x + c 3 4i eix + c 4 e ix 18

19 Let c 1 5 and c 5 = c and c 3 4 = c 3 the above simplifies to Convert to trig using Euler s we obtain y h e x + c e x e ix c 3 + c 4 e ix i e x + c e x + c 3 ie ix + c 4 e ix y h e x + c e x + c 3 i cos x + i sin x) + c 4 cos x i sin x) e x + c e x + cos x) ic 3 + c 4 ) + sin x) c 3 ic 4 ) Let ic 3 + c 4 ) = c 5 and c 3 ic 4 ) = c 6, new constants, hence Finding the particular solutions y h e x + c e x + c 5 cos x) + c 6 sin x) To find the particular solution, using superposition. Since D 4 + 3D 4) y = sinh x) sin x, we solve first for the first forcing function D 4 + 3D 4 ) y = sinh x) sinh x) can not be used for trail solution, as the homogeneous solution include e x in it and sinh x) = e x + ex. Therefore we will use Axex + Cxe x as trial solution. Hence y p1 = Ae x + Bxe x + Ce x + Dxe x y p1 = Ae x + Be x + Bxe x Ce x + De x Dxe x y p1 = Ae x + Be x + Be x + Bxe x + Ce x De x De x + Dxe x = Ae x + Be x + Bxe x + Ce x De x + Dxe x y p1 = Ae x + Be x + Be x + Bxe x Ce x + De x + De x Dxe x = Ae x + 3Be x + Bxe x Ce x + 3De x Dxe x y p1 = Ae x + 3Be x + Be x + Bxe x + Ce x 3De x De x + Dxe x = Ae x + 4Be x + Bxe x + Ce x 4De x + Dxe x Therefore the ODE becomes, and using ex e x for sinh x) D 4 + 3D 4 ) y p1 = Ae x + 4Be x + Bxe x + Ce x 4De x + Dxe x) + 3 Ae x + Be x + Bxe x + Ce x De x + Dxe x) Hence, comparing coefficients 4 Ae x + Bxe x + Ce x + Dxe x) = ex e x e x A + 4B + 3A + 6B 4A)+e x C 4D + 3C 6D 4C)+xe x B + 3B 4B)+xe x D + 3D 4D) = ex Hence A + 4B + 3A + 6B 4A = 1 C 4D + 3C 6D 4C = 1 19

20 Hence 10B = 1 10D = 1 Hence B = 1 0 D = 1 0 Therefore y 1p = 1 0 xex xe x To find second particular solution, D 4 + 3D 4 ) y = sin x Since sin x = e ix + e ix ) and the functions e ±ix are in the homogeneous solution, let the trial function be y p = a + bxe ix + cxe ix. Plug-in this into the ODE and expanding gives e ix 3ib + 16xb 1ib 4bx) + e ix 3ic + 16xc + 1ic 1cx 4bx) 4a = e ix + e ix) This can be used to find y p need to more time to work this out). The final solution will then be y = y h + y p1 + y p y h e x + c e x + c 5 cos x) + c 6 sin x) xex xe x + y p note: I verified the solution using Mathematica. The homogeneous solution appears to be correct, but need to work more on the particular solution. Here is the result Where y p was given as e x where y x) e x + c e x + c 5 cos x) + c 6 sin x) + y p = 80e x 0e x + 40e x x + 40x 0e x sin x) + 0e x x sinx) + 5e x sinx) sin4x) + 10e x cos 3 x) 0e x cos x)+ 16e x cosx) 3e x sin x) sinhx) 3e x cos x) sinhx) + 0 I tried using the variational method, but needed more time to complete finding the particular solution. 0