What makes a differential equation linear as opposed to non-linear?
Linear equations will have y terms either by themselves or multiplied by a function of x. You will never see y itself being manipulated (i.e. y^2, (1-y)y')
What does it mean for a differential equation to be autonomous?
The independent variable does not appear explicitly.
What is a critical point of a function?
A point where the derivative is zero.
What does a phase portrait look like?
A phase portrait is a simplified version of a slope field diagram. There is simply a single arrow along the y axis in each region, indicating whether the derivative in that region is positive or negative.
What makes a linear equation of the form a(dy/dx) + b(x)y = g(x) homogeneous?
The equation is homogeneous if g(x) = 0.
How do you show that an equation of the form M(x,y) dx + N(x,y) dy = 0 is exact?
By showing that the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x.
What equation form would use an integrating factor?
dy/dx + P(x)y = f(x)
What is the formula for an integrating factor?
I(x) = e^(∫P(x) dx)
Which equation format would be solved using an integrating factor and how would you apply the integrating factor?
- dy/dx + P(x)y = f(x)
- d/dx((y)I(x)) = f(x)I(x)
What does it mean for an equation of the form M(x,y) dx + N(x,y)dy = 0 to be homogeneous?
The function is homogeneous if M and N are homogeneous functions of the same degree (alternative form: if M(tx, ty) = t^aM(x,y) and N(tx, ty) = t^aN(x,y)).
What is the format of a Bernoulli equation?
dy/dx + P(x)y = f(x)y^n
How does the format of a Bernoulli equation differ from the format dy/dx + P(x)y = f(x)?
Bernoulli equations involve f(x) being multiplied by a power of y.
How do you solve a Bernoulli equation?
- dy/dx + P(x)y = f(x)y^n
- Set u = y^(1-n)
- Solve for du and substitute in u and du.
- Solve the equation, finishing by substituting back in y
How can you solve and equation of the form dy/dx = f(Ax + By + C)?
- Set u = (Ax + By + C) and solve the equation using u.
- For example, dy/dx = (-2x + y)^2 - 7 would have a substitution of u = (-2x + y).
What is generally the best method to solve a logistic equation: dP/dt = P(a - bP)?
Usually partial fraction decomposition works well.
Show that (2x + 5y) dx + (5x - y^2) dy = 0 is exact.
- partial derivative of M with respect to y = 5
- partial derivative of N with respect to x = 5
- The two partial derivatives are equal, so the solution is exact.
What is the best method to solve (2x + 5y) dx + (5x - y^2) dy = 0 and why? Demonstrate how this method works.
- Because the differential equation is exact, the best method is to find the original function by integrating each component by the derivative they are being multiplied by, and then resolving the equations.
- f(x) = x^2 + 5xy + g(y)
- g(y) = 5xy - 1/3y^3 + f(x)
- F(x,y) = x^2 + 5xy - 1/3y^3 + C
What is the one thing that you're always going to have to remind yourself of when integrating?
The damn +c
How would you show that (x-y)dx + x dy = 0 is homogeneous?
- M and N are both homogeneous, or
- (A^z)M(x,y) = M(Ax, Ay) and (A^z)N(x,y) = N(Ax, Ay)
- 5(x-y) = 5x - 5y 5(x) = 5x
- 5x - 5y = 5x - 5y 5x = 5x
How would you solve (x-y) dx + x dy = 0 and why?
- The equation is not exact, nor is it easily separable, so you would use a u substitution.
- Set y = ux. Then dy = u dx + x du
- (x - ux) dx + x(u dx + x du) = 0
- (x - ux) dx + ux dx + x^2 du = 0
- x^(-1) dx + du = 0
- ln(x) + u = C
- ln(x) + y/x = C
- y/x = C - ln(x)
- y = xC - xln(x)
Determine the order and linearity of d2y/dx2 + (sin x)y = cos x.
Order = 2, Linearity = Linear
Why is (dy/dx)^2 + dy/dx + x^2y^2 non-linear?
In the first term, ther derivative is squared. Additionally, in the third term, y is multiplied by itself instead of just a function of x.
How would you determine if y = e^(3t) is a solution to y'' - 9y = 0?
- Solve for y'' and plug the two values into the differential equation.
- y' = 3e^(3t)
- y'' = 9e^(3t)
- 9e^(3t) - 9(e^3t) = 0
- 0 = 0
How would you solve dy/dx = e^x*e^(-2y) and why?
- This equation is separable. Therefore you would group the x's and y's, respectively, and integrate to solve.
- e^2y dy = e^x dx
- (1/2)e^2y = e^x + C
- e^2y = 2e^x + C1
- ln(e^2y) = ln(2*e^x) + ln(C1)
- 2y = ln(2) + x + C2
- y = 2ln(2) + 2x + C3
How would you solve dy/dx + 6y = e^(4x) and why?
- This equation is in the form dy/dx + P(x)y = f(x), so you would use an integrating factor
- I(x) = e^(∫P(x) dx)
- d/dx(y*I(x)) = f(x)I(x)
- I(x) = e^(∫6) dx
- I(x) = e^6x
- d/dx(y*e^6x) = e^4x*e^6x
- y*e^6x = (1/10)e^10x + C
- y = (1/10)e^4x + Ce^(-6x)
What is the one thing you should never forget when integrating?
The damn +C
Where are the inflection points on a graph?
The points where the second derivative changes sign, or where the graph goes from concave up to concave down or vice versa.
What solution method can be used to solve an equation of the form dx/dt = k(a - (m/(m+n))x)(b - (n/(m+n))x)?
- This can be solved by using Bernoulli:
- dx/dt + (b-(n/(m+n))x = (a - (m/(m+n))x
Set up a system of differential equations for this situation.
- dx1/dt = rin - rout
- dx1/dt = (0 + (x2/50)1) - ((x1/50)4)
- dx2/dt = rin - routdx2/dt = ((x1/50)4) - ((x2/50)(3 + 1))
- dx1/dt = x2/50 - 2x1/25
- dx2/dt = 2x1/25 - 2x2/25
Set up a system of equations for this situation in terms of i2
- E = L(di/dt) + Ri
- E(t) = R1i1 + L1(di2/dt) + R2i3E(t) = R1(i2 + i3) + R2i3 + L1(di1/dt) = i2R1 + i3(R1 + R2) + L1(di2/dt)
- E(t) = R1i1 + L2(di3/dt) = R1(i2 + i3) + L2(di3/dt)
What do you need to remember in every integration?
The damn plus c