Y2, C7 - Differential Equations Flashcards
How would you separate the variables in: dy/dx = f(x) g(y)
1/g(y) * dy/dx = f(x)
int(1/g(y)) * dy = int(f(x)) * dx
Find general solutions to dy/dx = -y/x
1/y * dy/dx = -1/x
int(1/y) * dy = int(-1/x) * dx
lnlyl = -lnlxl + lnlkl
y = k/x
Solve d/dx * (ysin(x))
sin * dy/dx + ycos(x)
Find general solutions of the equation x^3 dy/dx + 3x^2 * y = sinx
d/dx * (x^3 * y) = sinx
x^3 * y = int(sinx) dx
x^3 * y = -cosx
x^3 * y = -cos(x) + c
Using the reverse product rule, find general solutions to 4xy * dy/dx + 2y^2 = x^2
d/dx * (2xy^2) = x^2
2xy^2 = int(x^2)dx
2xy^2 = (1/3) * x^3 + c
y^2 = (1/6) * x^2 + c/2x
What is the integrating factor
e^(int(P)dx)
How do you solve equations in the form dy/dx + Py = Q
(where P and Q are functions of x)
Reverse product rule
What form of equation can you use the reverse product rule on
dy/dx + Py + Q
(where P and Q are functions of x)
What is the integrating factor of dy/dx - 4y = e^x
I.F. = e^(int(-4dx))
= e^-4x
What do you do once you have found the integrating factor
Multiply all terms in the expression by the integrating factor
Then solve the usual way (reverse product rule)
What do you do if there is a coefficient in front of the dy/dx term in an expression of form dy/dx + Py = Q
Divide all terms by the coefficient to remove
What would the solution of a * dy/dx + by = 0 be in the form of
y = Ae^(x * -b/a)
What would the solution of a(d2y/dx2) + b(dy/dx) + cy = 0 be in the form of
y = Ae^mx
What is the Auxiliary equation
am^2 + bm + c = 0
What does auxiliary mean
Helpful
What is the general solution to a differential equation if the auxiliary equation has two distinct roots
y = Ae^αx + Be^βx
What is the auxiliary equation of 2(d2y/dx2) + 5(dy/dx) + 3y = 0
2m^2 + 5m + 3 = 0
When the auxiliary equation has two equal roots, what is the general solution
y = (A + Bx)e^αx
When the auxiliary equation has no real roots, what is the general equation
y = Ae^αx + Be^βx
(the same as if the roots are real)
Find the general solution of d2y/dx2 + 16y = 0
y = Ae^4ix + Be^-4ix
= A(cos4x + isin4x) + B(cos4x - isin4x)
= (A+B)cos4x + (A-B)isin4x
y = Pcos4x + Qsin4x
P = A+B
Q = (A-B)i
If the auxiliary equation has two imaginary roots ±Ω, what is the general solution
y = Acos(Ωx) + Bsin(Ωx)
where A and B are arbitrary constants
If the auxiliary equation has two complex roots p ±iq, what is the general solution
y = e^px (Acosqx + Bsinqx)
where A and B are arbitrary constants
Find the general solution to d2y/dx2 - 6 * dy/dx + 34y = 0
y = Ae^(3+5i)x + Be^(3-5i)x
y = e^3x * (Pcos5x + Qsin5x)
What do you do if you have a non-homogeneous second order differential equation = f(x) (not equal to zero on RHS)
Solve as if RHS = 0 to obtain ‘complimentary function’ (C.F.
Then solve equation = f(x), can be found by using appropriate substitutions and comparing coefficients (solution known as particular integral (P.I.))
y = C.F. + P.I. as C.F. = 0 and P.I. = f(x) which sum to f(x)
