Dif Eq

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')

The independent variable does not appear explicitly.

A point where the derivative is zero.

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.

The equation is homogeneous if g(x) = 0.

By showing that the partial derivative of M with respect to y is equal to the partial derivative of N with respect to x.

dy/dx + P(x)y = f(x)

I(x) = e^(∫P(x) dx)

• dy/dx + P(x)y = f(x)
• d/dx((y)I(x)) = f(x)I(x)

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)).

dy/dx + P(x)y = f(x)y^n

Bernoulli equations involve f(x) being multiplied by a power of y.

• 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

• 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).

Usually partial fraction decomposition works well.

• 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.

• 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

The damn +c

• 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

• 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)

Order = 2, Linearity = 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.

• 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

• 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

• 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)

The damn +C

The points where the second derivative changes sign, or where the graph goes from concave up to concave down or vice versa.

• This can be solved by using Bernoulli:
• dx/dt + (b-(n/(m+n))x = (a - (m/(m+n))x

• dx₁/dt = r_in - r_out
• dx₁/dt = (0 + (x₂/50)1) - ((x₁/50)4)
• dx₂/dt = r_in - r_out
• dx₂/dt = ((x₁/50)4) - ((x₂/50)(3 + 1))
• dx₁/dt = x₂/50 - 2x₁/25
• dx₂/dt = 2x₁/25 - 2x₂/25

• E = L(di/dt) + Ri
• E(t) = R₁i₁ + L₁(di₂/dt) + R₂i₃
• E(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)

The damn plus c