A Differential Equation (DE) is an equation that relates a function with its derivatives.
This unit focuses on First Order, First Degree equations, which have the general form dy/dx = f(x, y) or M(x, y)dx + N(x, y)dy = 0.
Standard types and their solution methods:
The equation can be algebraically rearranged into the form f(y) dy = g(x) dx.
Solution: Integrate both sides: ∫ f(y) dy = ∫ g(x) dx + C
The equation can be written in the form dy/dx = F(y/x).
Solution: 1. Substitute y = vx. 2. This implies dy/dx = v + x(dv/dx). 3. The equation becomes v + x(dv/dx) = F(v). 4. This is now separable: x(dv/dx) = F(v) - v => dv / (F(v) - v) = dx / x. 5. Integrate and then substitute v = y/x back.
The equation has the standard form: dy/dx + P(x) · y = Q(x).
Solution: 1. Find the Integrating Factor (IF):
This equation has the form: dy/dx + P(x) · y = Q(x) · yn.
Solution: 1. Divide the entire equation by yn: y-n(dy/dx) + P(x)y(1-n) = Q(x). 2. Make the substitution: z = y(1-n). 3. Then dz/dx = (1-n)y-n(dy/dx). 4. Substitute these into the DE. It will become a linear DE in 'z' and 'x', which you can solve using the IF method. 5. Substitute z = y(1-n) back at the end.
An equation of the form M(x, y)dx + N(x, y)dy = 0 is "exact" if it is the total differential of some function u(x, y).
i.e., M = ∂u/∂x and N = ∂u/∂y.
Condition for Exactness:
The equation is exact if and only if:∂M/∂y = ∂N/∂x
If the equation is exact, the general solution is:
If a DE of the form Mdx + Ndy = 0 is not exact (i.e., ∂M/∂y ≠ ∂N/∂x), we can sometimes find an Integrating Factor (IF). This is a function we multiply the entire equation by to *make* it exact.
Rule 1: If [ (∂M/∂y) - (∂N/∂x) ] / N = f(x) (a function of x alone)
Rule 2: If [ (∂N/∂x) - (∂M/∂y) ] / M = g(y) (a function of y alone)
Rule 3: If the equation is homogeneous and (Mx + Ny) ≠ 0
Rule 4: If the equation is of the form y·f(xy)dx + x·g(xy)dy = 0 and (Mx - Ny) ≠ 0
This is a general technique where a substitution is used to transform a complicated DE into a simpler, standard form (like separable or linear).
General Strategy: 1. Look for a repeated group in the equation, e.g., (x + y). 2. Substitute this group: z = x + y. 3. Find the derivative: dz/dx = 1 + dy/dx => dy/dx = (dz/dx) - 1. 4. Substitute 'z' and 'dy/dx' into the original equation. 5. The new equation in 'z' and 'x' should be simpler (e.g., variables separable). 6. Solve for 'z', then substitute back z = x + y.
This method is highly flexible and depends on spotting the right substitution.
This is an introduction to the equations that will be solved in Unit 5. A linear DE of order 'n' has the general form:
Unit 5 will focus on solving these equations, particularly when the coefficients are constants.