Topic 10: Differential Equations

Question 1

What a first order separable differentiable equations, and how do you solve them (1,2)

Answer 1

-These are of the form dy/dx = f(x) g(y)

-You first rearrange these, to get 1/g(y) dy/dx = f(X)
-You then integrate as ∫1/g(y) (dy/dx) dx = ∫1/g(y) dy = ∫f(x) dx + c

Question 2

How can we solve dy/dx = xy (5)

Answer 2

dx/dy = xy
-(1/y) dy/dx = x
-∫(1/y)dy = ∫x dx
-lny = (1/2)x² + c
-y = e^{(1/2)x2 + c}
-Let A = e^c, so that y = Ae^(1/2)x2

Question 3

What are first order linear equations (1)

Answer 3

-dy/dx + a(x)y = b(x)

Question 4

How can we rearrange x²dy/dx + 2xy = b(x) to help us solve the first order differential equation (5,1)

Answer 4

-x²dy/dx + 2xy = (d/dx)(x²y)
-∫ (d/dx)(x²y) dx= ∫ b(x) dx
=x²y = ∫ b(x) dx
-y = (1/x²) ∫ b(x) dx
-y = (1/x²) ∫ b(x) dx + c/x²

-Solve the integral to get the general solution, then substitute in the initial/boundary conditions to find c for the required specific solution

Question 5

How can we let u(x) = e^{∫ a(x) dx} to help find the integrating factor (6,1)

Answer 5

-Let u(x) = e^{∫ a(x) dx}
-du/dx = a(x)u(x)
-Given dy/dx + a(x)y = b(x), we get:
-u(x)dy/dx + a(x)u(x)y = u(x)b(x)
-This is equal to (d/dx)(u(x)y) = u(x)b(x)
-Then we can integrate dx

-We can ignore the constant of integration when calculating u(x) = e^{∫ a(x) dx}

Question 6

How can we solve the first-order differential equation dy/dx - y = x (6)

Answer 6

-Find u(x) = e^{∫-1 dx} = e^-x
-So then e^-x(dy/dx) - ye^-x = xe^-x
-(d/dx)(e^-xy) = xe^-x (use integration by parts for the RHS)
-e^-xy = -xe^-x - e^-x + c = -e^-x(1+x) + c
-y = -(1+x) + ce^x
-This is the general solution, to find the specific solution cconsider various solution curves

Question 7

What is the integrating factor (2)

Answer 7

-The integrating factor is used with differential functions in the form dy/dt + a(t)y = b(t)
-The end goal is getting the LHS in form d/dt(uv) = u(dv/dt) + v(du/dt)

Question 8

How do you calculate the integrating factor (1,1,1,6,3)

Answer 8

-Start with the equation in form dy/dt + a(t)y = b(t)

-Evaluate ∫a(t) dt (no arbitrary constant needed)

-Determine u(t) = e^{∫a(t) dt} (this is the integrating factor)

-Multiply the differential equation through by u(t)
-dy/dt (u(t)) + (u(t))(a(t))y = (u(t))b(t)
-Since u(t) = e^{∫a(t) dt}, du/dt = a(t) e^{∫a(t) dt} = a(t) u(t)
-d/dt(u(t)y) = (dy/dt)(u(t)) + (du/dt)(y)
-d/dt(u(t)y) = (dy.dt)(u(t) + a(t)u(t)y
-d/dt(u(t)y) = u(t)b(t)

-We then integrate this
-u(t)y = ∫u(t)b(t) dx + c
-y = (1/u(t))(∫u(t)b(t) dx + c)

Question 9

How can we solve the linear equation dx/dt - 2tx = t (4)

Answer 9

-The integrating factor is u(t) = e^{∫-2t dt} = e^-t2
-The differential equation then becomes (d/dt)(e^-t2x) = te^-t2)
-e^-t2x = ∫te^-t2 dt + c = (-1/2)e^-t2 + c (integrate with respect to dt)
-x = Ce^t2 - 1/2 (this is the general solution)

Question 10

What are second order differential equations (1)

Answer 10

-(d²y/dx²) = F(x, y, dy/dx) or x̄ = F(t, x, x’)

Question 11

How do we directly integrate second order differential equations, and what information do we need about x (3,3)

Answer 11

-take d²x/dt² = k
-dx/dt = kt + c
-x = (1/2)kt² + Ct + D

To determine C & D, we need 2 pieces of information about x:
-2 points through which the solution curve passes (boundary value problem)
-x and (dx/dt) at the same point (initial value problem)
-x at one point, dx/dt at another (mixed problem)

Question 12

How can we solve d²x/dt² = dx/dt + t, given x(0) = 1 and ẋ(0) = 2 (7,1)

Answer 12

-Replace dx/dt with new variable y, to make the equation a first order one
-ẏ = y + t, y(0) = 2
-y = -(1+t) + Ce^t
-When t = 0, y = 2 so c = 3
-ẋ = -(1+t) + 3e^t
-x = -t -(1/2)t² + 3e^t + D
-x(0) = 1, D = -2

-x = -(2+t+(1/2)t²) + 3e^t

Question 13

What is the form of linear second order equations, and when are they homogeneous vs inhomogeneous (3,2)

Answer 13

-a(t) d²y/dt² + b(t)dy/dt + c(t)y = f(t), where a(t), b(t), c(t) f(t) are continuous functions of t on some interval a(t) ≠ 0
-Ly = f(t)
-L = the linear differential operator, a(t) d²/dt² + b(t)d/dt + c(t)

-If f(t) = 0, the equation is said to be homogeneous
-if f(t) ≠ 0, the equation is inhomogeneous

Question 14

How can we combine solutions of linear homogeneous second-order differential equations (1,2)

Answer 14

-Let V be the set of all solutions, and let u₁, u₂ ∈ ℝ such that Lu₁, Lu₂ = 0

-u₁ + u₂ is a solution
-ku is a solution

Question 15

What do we call V when solving linear homogeneous second-order differential equations (2,2)

Answer 15

-V is the vector space consisting of all the functions which are mapped to the 0 function by the linear map L
-We can combine the 2 properties and say Au₁ + Bu₂ is a solution for all choices of real arbitrary constants A and B

However, the general solution to the equations is in the form Au₁ + Bu₂ only when:
-u₁ and u₂ span the solution space of Ly
-u₁ and u₂ are linearly independent

Question 16

How do we solve linear non-homogeneous second-order differential equations (3)

Answer 16

The non-homogeneous differential equation has the general solution:
-y = Au₁(t) + Bu₂(t) + v(t)
-Au₁(t) + Bu₂(t) is called the complementary solution, and is the general solution of the related homogeneous equation
-v(t) is any particular solution so that Lv = f(t), provided v ∉ V

Question 17

How can we write the linear differential operator for linear non-homogeneous second-order differential equations (3)

Answer 17

-L(Au₁ + Bu₂ + v) = AL(u₁) + BL(u₂) + Lv
=0 + 0 + Lv
-f(t) = Lv

Question 18

How can we find the auxillary or characteristic equation of a second-order linear differential equation with constant coefficients (1,6)

Answer 18

-a(d²y/dt²) + b(dy/dt) + cy = 0

-If we try solution y = e^mt, dy/dt = me^mt, d²y/dt² = m²e^mt
-a(m²e^mt) + b(me^mt) + c(e^mt) = 0
-e^mt(am² + bm + c) = 0
-Since e^mt ≠ 0, am² + bm + c = 0
-This is an auxillary or characteristic equation
-This has solutions m₁, m₂ = (-b±√(b² - 4ac))/2a

Question 19

How can we find the complementary solution to a second-order linear equation with constant coefficients (3,4,3,5)

Answer 19

-a(d²y/dt²) + b(dy/dt) + cy = 0
-e^mt(am² + bm + c) = 0 (let y = e^mt)
-This has solutions m₁, m₂ = (-b±√(b² - 4ac))/2a

-Say m₁ and m₂ are real and distinct (b² > 4ac)
-u₁ = e^m₁t, u₂ = e^m₂t
-These are linearly independent functions, so the complementary solution is a linear combination
-Ae^m₁t + Be^m₂t

-Say m₁ = m₂ =m (b² = 4ac)
-u₁ = e^mt, u₂ = te^mt
-Complementary solution: (A + Bt)(e^mt)

-Say m₁ and m₂ = α + iβ (b² < 4ac)
-u₁ = e^{(α + iβ)t} = e^αt(cosβt + i(sinβt))
-u₂ = e^{(α - iβ)t} = e^αt(cosβt - i(sinβt))
-Since these are not real, we choose 2 real linearly independent solutions from the complex subspace spanned by u₁ and u₂ ((u₁ and u₂)/2)
-General solution: e^αt(Acosβt + Bsinβt)

Question 20

How can we solve (d²x/dt²) + 4(dx/dt) + 4x = 0 (3)

Answer 20

-The characteristic equation is m² + 4m + 4 = 0
-Solving this gives us m = -2
-The general solution is thus x = (A + Bt)e^-2t

Question 21

How can we solve (d²y/dt²) -6 (dy/dt) + 13y = 0

Answer 21

-The characteristic equation is m² - 6m + 13
-Using the quadratic equation, we get 3 ± √-1 = 3 ± 2i
-The general solution is thus y = e^3t(A cos(2t) + B sin(2t))

Question 22

What is the particular solution, and how do we find it (1,1)

Answer 22

-The particular solution is the bit to add to the complementary to get the general solution to the original non homogeneous equation

-We first use a trial and error solution, then change the coefficients

Question 23

What should our trial solutions for particular solutions be depending on the original nonhomogeneous equations (1,7,1)

Answer 23

-Remember a(t) d²y/dt² + b(t)dy/dt + c(t)y = f(t)

-If f(t) is a constant, trial C
-If f(t) = t, trial Ct + D
-If f(t) = t², trial ct² _ Dt + e
-If f(t) = tⁿ, trial a polynomial of the same degree
-If f(t) = e^kt, trial Ce^kt
-If f(t) = sin(at), trial C(cos(at)) + D(sin(at))
-If f(t) = cos(at), trial C(cos(at)) + D(sin(at))

-However, since the trial solution has to be linearly independent from the complementary, for f(t) = e^kt be prepared to use Cte^kt or Ct²e^kt

Question 24

How can we solve (d²x/dt²) +5 (dx/dt) + 6x = 4 (1,1,3,1)

Answer 24

-The auxillary solution is m² + 5m + 6 = 0, meaning m = -2, -3

-The complementary solution is thus Ae^-2t + Be^-3t

The particular solution is
-f(t) = 4
-Let x = c, so dx/dt = d²x/dt² = 0
-This gives us 6C = 4, so C = 2/3

-The general solution is thus 2/3 + Ae^-2t + Be^-3t

Question 25

How can we solve (d²x/dt²) +5 (dx/dt) + 6x = 3e^2t (1,1,3,1)

Answer 25

-The auxillary solution is m² + 5m + 6 = 0, meaning m = -2, -3 -The complementary solution is thus Ae^-2t + Be^-3t The particular solution is: -f(t) = 3e^2t -Try x = Ce^2t, dx/dt = 2Ce^2t, d²x/dt² = 4Ce^2t -Subbing into the differential equation gives us 20C(e^2t) = 3e^2t, C = 3/20 -The general solution is hence (3/20)e^2t + Ae^-2t + Be^-3t

Question 26

What is a summary on how to solve a(d²y/dx²) + b(dy/dx) + cy = v(x) (5)

Answer 26

-Solve the characteristic equation am² + bm + c = 0 -Write down the complementary solution u: Ae^m₁x + Be^m_2/sub>x, (A+Bx)e^mx or e^px(Acos(qx) + Bsin(qx)) -Find the particular solution v(x) -y = u + v -Now find the specific solution by subbing the boundary conditions into the general solution and solving the resulting simultaneous equations