Home > Preview
The flashcards below were created by user
klockhart
on FreezingBlue Flashcards.

What makes a differential equation linear as opposed to nonlinear?
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, (1y)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^(1n)
 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 (xy)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(xy) = 5x  5y 5(x) = 5x
 5x  5y = 5x  5y 5x = 5x

How would you solve (xy) 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 nonlinear?
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.
 dx_{1}/dt = r_{in}  r_{out}
 dx_{1}/dt = (0 + (x_{2}/50)1)  ((x_{1}/50)4)
 dx_{2}/dt = r_{in } r_{out}
 dx_{2}/dt = ((x_{1}/50)4)  ((x_{2}/50)(3 + 1))
 dx_{1}/dt = x_{2}/50  2x_{1}/25
 dx_{2}/dt = 2x_{1}/25  2x_{2}/25

Set up a system of equations for this situation in terms of i _{2 }and i _{3}.
 E = L(di/dt) + Ri
 E(t) = R_{1}i_{1} + L_{1}(di_{2}/dt) + R_{2}i_{3}
 E(t) = R_{1}(i_{2 }+ i_{3}) + R_{2}i_{3} + L_{1}(di_{1}/dt) = i_{2}R_{1} + i_{3}(R_{1} + R_{2}) + L_{1}(di_{2}/dt)
 E(t) = R_{1}i_{1} + L_{2}(di_{3}/dt) = R_{1}(i_{2} + i_{3}) + L_{2}(di_{3}/dt)

What do you need to remember in every integration?
The damn plus c

You have an equation in the form M dx + N dy = 0. What is the first thing you do and why?
Check for exactness. If the equation is not exact, you can use the f(x) and g(y) approach.

What is wrong with this problem?:
E(t) = L di/dt = Ri
12 = .2 di/dt + 20i
.2 di/dt + 20i = 12
I(x) = e^{∫P(x) dx} = e^{∫20 dt} = e^{20t}
d/dt(ie^{20t}) = 12e^{20t}
ie^{20t} = .6e^{20t} + C
i = .6 + Ce^{20t}
.5 = .6 + C
C = .1
i = .6  .1e^{20t}
 In the fourth row, the integrating factor is taken while the form of the equation is k(arbitrary function or constant) dy/dx + P(x)y = f(x), rather than dy/dx + P(x)y = f(x). The equation needs to be divided by the coefficient of di/dt before the integrating factor can be taken.
 E(t) = L di/dt = Ri
 12 = .2 di/dt + 20i
 .2 di/dt + 20i = 12
 di/dt + 100i = 60
 I(x) = e^{∫P(x) dx} = e^{∫100dt }= e^{100dt}
 d/dt(ie^{20t}) = 12e^{20t}
 ie^{20t} = .6e^{20t} + C
 i = .6 + Ce^{20t}
 .5 = .6 + C
 C = .1
 i = .6  .1e^{20t}

Other than the +C, what is one thing you always need to check for when solving a dif. eq.?
Initial conditions to solve for the +C

What is one thing you need to be careful about when taking the integral of any equation?
Make sure you're taking the integral and not the derivative.

Solve the DE: dy/dx + y^{2}/(1 + x^{2}) = 0.
 dy/dx = y^{2}/(1 + x^{2})
 y^{2} dy = 1/(x^{2} + 1) dx
 y^{1} = tan^{1}(x) + C
 y^{ }= (tan^{1}(x) + C)^{1}

Solve the DE: y' + 3x^{2}y = x^{2}
 I(x) = e^{∫P(x) dx} = e^{∫3x^2 dx} = e^{x^3}
 d/dx(ye^{x^3}) = x^{2}e^{x^3}
 ye^{x^3} = 1/3e^{x^3} + C
 y = 1/3 + Ce^{x^3}

Solve the DE and give the largest intervals on which a solution can exist: xy' + y = e^{2x}
 dy/dx + x^{1}y = x^{1}e^{2x}
 I(x) = e^{∫x^(1) dx} = e^{ln x} = x
 d/dx(yx) = x(x^{1})e^{2x}
 d/dx(yx) = e^{2x}
 yx = .5e^{2x} + C
 y = x^{1}(.5^{}e^{2x} + C)
 Solution exists (∞, 0) (0, ∞)

A tank initially contains 300 gallons of saline solution with salinity 5 lbs of salt per gallon of solution. solution having salinity of 4 lbs of salt per gallon is pumped into the tank at a rate of 8 gallons per minute, and solution leaves the tank at the same rate of 8 gallons per minute. Fine the function A(t) which models the amount of salt in the tank at any time, and find the amount of salt in the tank after a long time. Be sure to draw a picture.
 dA/dt = r_{in } r_{out}
 dA/dt = (4*8)  8(A(t)/300)
 dA/dt = 32  2A(t)/75
 dA/dt + 2A(t)/75 = 32
 I(x) = e^{∫P(x) dx} = e^{∫2/75} = e^{2/75t}
 d/dt(A(t)e^{2/75t} = 32e^{2/75t}
 A(t)e^{2/75t} = 1200e^{2/75t} + C
 A(t) = 1200 + Ce^{2/75t}
 300*5 = 1200 + C
 C = 300
 A(t) = 1200 + 300e^{2/75t}
 300*4 = 1200
 After a long time there will be 1200 lbs of salt in the tank.

5. Show that the differential form below is not exact. Multiply both side by the integrating factor 1 + x and verify that the resulting differential form is exact. Solve the DE.
(x^{2} + y^{2}) dx + (y + xy) dy = 0
 ∂M/∂y = 2xy
 ∂N/∂x = y
 2xy != y, so the equation is not exact.
 (1 + x)(x^{2} + y^{2}) dx + (1 + x)(y + xy) dy = 0
 (x^{2} + y^{2 }+ x^{3} + xy^{2}) dx + (y + xy + xy + x^{2}y) dy = 0
 (x^{3} + x^{2} + xy^{2} + y^{2}) dx + (x^{2}y + 2xy + y) dy = 0
 ∂M/∂y = 2xy + 2y
 ∂N/∂x = 2xy + 2y
 2xy + 2y = 2xy + 2y, so the equation is exact.
 f(x) = .25x^{4} + .33x^{3} + .5x^{2}y^{2 }+ xy^{2 }+ g(y)
 g(y) = .5x^{2}y^{2 }+ xy^{2} + .5y^{2} + f(x)
 F(x,y) = .25x^{4} + .5x^{2}y^{2} + .33x^{3} + xy^{2} + .5y^{2} + C

For a series circuit containing only a resistor and an inductor, Kirchoff's second law states that the sum of the voltage drop across the inductor (L(di/dt)) and the voltage drop across the resistor (iR) is equal to the impressed voltage (E(t)) on the circuit. Suppose that a 12volt battery is connected to an LR series circuit in which the inductance is .2 henry and the resistance is 20 ohms. Determine the current i if the initial current is .5amp.
 E(t) = L (di/dt) + Ri
 12 = .2 di/dt + 20i
 .2 di/dt + 20i = 12
 di/dt + 100i = 60
 I(x) = e^{∫P(x) dx} = e^{∫100 dt} = e^{100t}
 d/dt(ie^{100t}) = 60e^{100t}
 ie^{100t} = .6e^{100t} + C
 i = .6 + Ce^{100t}
 .5 = .6 + C
 C = .1
 i = .6 + .1e^{100t}

Solve the IVP:
(xy) dx + x dy = 0; y(1) = 1
 ∂M/∂y = 1
 ∂N/∂y = 1
 1 != 1; equation is not exact, so cannot use f(x), g(y).
 Instead use u substitution
 y = ux dy = u dx + x du
 (x  ux) dx + x (u dx + x du) = 0
 (x  ux) dx + xu dx + x^{2} du = 0
 x dx + x^{2} du = 0
 x^{1 }dx + du = 0
 ln(x) + u = C
 ln(x) + y/x = C
 y/x = ln(x) + C
 y = x(ln(x) + C)
 1 = 1(ln(1) + C)
 C = 1
 y = x(1  ln(x))

What are five different methods of solving differential equations?
Bernoulli (dy/dx + P(x)y = f(x)y^{n}), dy/dx + P(x)y = f(x), u substitution, F(x,y) = f(x) + g(y), utilizing the fact that an equation is separable

Solve the DE
dy/dx + y = e^{x}y^{2}
 u = y^{1n} = y^{1}; y = u^{1}; dy = u^{2 }du
 u^{2}du/dx + u^{1} = e^{x}u^{2}
 du/dx  u = e^{x}
 I(x) = e^{∫P(x) dx} = e^{∫1 dx} = e^{x}
 d/dx(ue^{x}) = e^{x} e^{x}
 d/dx(ue^{x}) = e^{2x}
 ue^{x} = .5e^{2x} + C
 u = .5e^{x} + Ce^{x}
 y^{1} = .5e^{x} + Ce^{x}
 y = (.5e^{x} + Ce^{x})^{1}

What are five type of differential equations/solution types to look out for?
Bernoulli (dy/dx + P(x)y = f(x)y^{n}), dy/dx + P(x)y = f(x), separable, F(x,y) = f(x) + g(y), u substitution

When an equation has a +C, what is one thing you always need to look for?
Initial conditions

