Elementary differential equations

Question 1

Fundamental theorem of calculus part (a)

Answer 1

(a) F(x) = ∫ (x0, x) f (= ∫ (x0, x) f(t) dt
defines a continuously differentiable function F on I, and F’ = f.
i.e. an antiderivative exists for any continuous function.

Question 2

Fundamental theorem of calculus part (b)

Answer 2

(b) Given a function f we have an antiderivative of F, such that F’ = f and ∫ (a, b) f = F(b) - F(a)

Question 3

What is the form of the first-order linear differential equation

Answer 3

from CB: u’(x) = f(x)u(x) + g(x) or u’(x) -f(x)u(x) = g(x)
from OS: y’ + p(x) = q(x)

Question 4

Steps for solving 1st order linear DE

Answer 4

1. Put equation in the standard form
2. Note the corresponding values for p(x) or f(x) and q(x) or g(x)
3. Determine the integrating factor I(x), I(x) = e^(∫p(x) dx)
4. Sub into equation to solve for u(x) or y where y = 1/I(x)[∫ I(x) q(x) dx + C]

Question 5

What is the form of a bernoulli differential equation

Answer 5

from CB: u’(x) + a(x)u(x) = b(x)u(x)ⁿ
from OS: Y’ + p(x)Y = q(x)Yⁿ

Question 6

Steps for solving bernoulli DE

Answer 6

1. Put equation into standard form
2. Note the corresponding values for p(x), q(x) and n
3. Determine the integrating factor I(x) such that I(x) = e^(∫(1-n)(p(x)) dx
4. Sub values into final equation for general solution:
   Y¹⁻ⁿ = 1/I(x) [∫ (1-n)(q(x)I(x))dx + c]
5. Try and manipulate the equation to get Y

Question 7

what is the form of separable DE

Answer 7

Y’ or dy/dx = p(x)q(y)

Question 8

Steps for solving separable DE

Answer 8

1. Put the equation into the standard form
2. Separate the x terms and the y terms
   (x-terms) dx = (y-terms) dy
3. integrate both sides
   ∫ (y-terms) dy = ∫ (x-terms) dx + C
4. Solve for Y

Question 9

What is the form of second order differential equations

Answer 9

P(x)y’’ + Q(x)y’ + R(x)y = G(x)
When G(x) = 0, the second order LDE is homogenous
when G(x) ≠ 0, the second order LDE is non-homogenous

Question 10

Steps for solving homogenous second order differential equations

Answer 10

1. When P(x), Q(x) and R(x) are continuous functions we have: ay’’ + by’ + cy = 0
2. We can take Y=e^rx so that y’=re^rx and y’’= r²e^rx so that
   ar²e^rx + bre^rx + ce^rx = 0 if we divide by e^rx we get the equation: ar² + br + c = 0
3. Sub in our constants a, b and c and solve for r.
4. Determine which equation you should use.
   If b² - 4ac > 0 there will be two roots and you should use equation Y = C₁e^r₁x + C₂e^r₂x
   If b² - 4ac = 0 there will be one real root where we solve for Y using equation Y = C₁e^rx + C₂xe^rx
   If b² - 4ac < 0 then there will be two imaginary roots where we solve for Y using equation Y = e^𝛼x [C₁cos(Bx) + C₂sin(Bx)