First-Order Differential Equations

Model real-world phenomena involving rates of change.

Introduction to Differential Equations

A differential equation is an equation that relates a function with its derivatives. These equations are fundamental in science and engineering as they describe how a quantity changes over time or space.

A first-order differential equation involves only the first derivative of the function.

Method: Separation of Variables

One of the simplest methods for solving first-order differential equations is the separation of variables. This method applies when you can algebraically separate the equation so that all terms involving `y` are on one side and all terms involving `x` are on the other.

Example: Solve dy/dx = x/y

  1. Separate the variables:
    y dy = x dx
  2. Integrate both sides:
    ∫ y dy = ∫ x dx
  3. Perform the integration:
    (y²)/2 = (x²)/2 + C₁
  4. Solve for y (General Solution):
    y² = x² + 2C₁
    Let C = 2C₁, so y² - x² = C. This represents a family of hyperbolas.

Method: Homogeneous Differential Equations

A differential equation of the form dy/dx = F(x, y) is homogeneous if F(x, y) can be expressed as a function of y/x alone. A common approach is to use the substitution y = vx.

Example: Solve dy/dx = (x+y)/x

  1. Check for homogeneity:
    dy/dx = 1 + y/x. This is a function of y/x.
  2. Substitute y = vx:
    This implies dy/dx = v + x(dv/dx).
  3. Rewrite the equation:
    v + x(dv/dx) = 1 + v.
  4. Separate variables:
    x(dv/dx) = 1, which separates to dv = dx/x.
  5. Integrate and substitute back:
    ∫ dv = ∫ dx/x => v = ln|x| + C.
    Since v = y/x, the solution is y/x = ln|x| + C, or y = x(ln|x| + C).

Method: Reducible to Homogeneous Form

Equations of the form dy/dx = (ax+by+c)/(Ax+By+C) can be reduced to homogeneous form by a shift of origin. Let x = X+h and y = Y+k.

Example: Solve dy/dx = (x-y+1)/(x+y-3)

  1. Find the new origin (h, k):
    Solve h-k+1=0 and h+k-3=0. Adding them gives 2h-2=0 => h=1. Then k=2.
  2. Substitute x = X+1, y = Y+2:
    dy/dx = dY/dX. The equation becomes dY/dX = (X-Y)/(X+Y), which is homogeneous.
  3. Solve the homogeneous equation:
    Using Y=vX, we get v + X(dv/dX) = (1-v)/(1+v).
  4. Separate and Integrate:
    X(dv/dX) = (1-v)/(1+v) - v = (1-2v-v²)/(1+v).
    ∫ (1+v)/(1-2v-v²) dX = ∫ dX/X. The left integral is of the form -1/2 ∫ du/u.
    This gives -1/2 ln|1-2v-v²| = ln|X| + C.
  5. Substitute back:
    Substitute v=Y/X, then X=x-1 and Y=y-2 to get the final implicit solution.

Method: Exact Differential Equations

An equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if ∂M/∂y = ∂N/∂x. The solution is ∫Mdx (treating y as constant) + ∫(terms in N not containing x)dy = C.

Example: Solve (2xy + 3)dx + (x² - 1)dy = 0

  1. Check for exactness:
    M = 2xy + 3, N = x² - 1.
    ∂M/∂y = 2x. ∂N/∂x = 2x. They are equal, so the equation is exact.
  2. Integrate M with respect to x:
    ∫(2xy + 3)dx = x²y + 3x.
  3. Integrate terms in N without x:
    The term in N without x is -1. ∫(-1)dy = -y.
  4. Combine for the solution:
    The general solution is x²y + 3x - y = C.

Method: Reducible to Exact Form

If an equation is not exact, it can sometimes be made exact by multiplying by an Integrating Factor (IF). If (∂M/∂y - ∂N/∂x)/N is a function of x alone, say f(x), then IF = e^(∫f(x)dx).

Example: Solve (x²+y²)dx - 2xydy = 0

  1. Check for exactness:
    M = x²+y², N = -2xy. ∂M/∂y = 2y, ∂N/∂x = -2y. Not exact.
  2. Find the Integrating Factor:
    (∂M/∂y - ∂N/∂x)/N = (2y - (-2y))/(-2xy) = 4y/(-2xy) = -2/x. This is a function of x.
    IF = e^(∫(-2/x)dx) = e^(-2lnx) = x⁻².
  3. Create the new exact equation:
    Multiply the original equation by x⁻²: (1+y²/x²)dx - (2y/x)dy = 0.
  4. Solve the new exact equation:
    New M = 1+y²/x², N = -2y/x. ∂M/∂y = 2y/x², ∂N/∂x = 2y/x². It's not exact. There is a calculation error.
    Let's check (∂N/∂x - ∂M/∂y)/M = (-2y - 2y)/(x²+y²) = -4y/(x²+y²). Not a function of y alone. There are other rules for IFs.

Method: Bernoulli's Form

A Bernoulli differential equation is of the form `y' + P(x)y = Q(x)yⁿ`. It can be reduced to a linear form with the substitution z = y^(1-n).

Example: Solve dy/dx + (1/x)y = x²y³

  1. Identify form:
    This is a Bernoulli equation with n=3.
  2. Substitute z = y^(1-3) = y⁻²:
    dz/dx = -2y⁻³(dy/dx). So dy/dx = (-y³/2)dz/dx.
  3. Rewrite the equation:
    (-y³/2)dz/dx + y/x = x²y³. Divide by -y³/2: dz/dx - 2/(xy²) = -2x².
    Since z = y⁻², this is dz/dx - (2/x)z = -2x². This is now a linear DE in z.
  4. Solve the linear DE:
    IF = e^(∫(-2/x)dx) = x⁻².
    Solution: z * x⁻² = ∫(-2x² * x⁻²)dx = ∫-2dx = -2x + C.
  5. Substitute back:
    y⁻² * x⁻² = -2x + C, or 1/(x²y²) = C - 2x.

Method: Reducible to Bernoulli's Form

Some equations can be transformed into Bernoulli's form through a suitable substitution.

Example: Solve dy/dx = y/(2y ln(y) + x)

  1. Invert the equation:
    dx/dy = (2y ln(y) + x)/y = 2ln(y) + x/y.
  2. Rearrange to linear form in x:
    dx/dy - (1/y)x = 2ln(y). This is a linear DE for x in terms of y.
  3. Find Integrating Factor:
    P(y) = -1/y. IF = e^(∫(-1/y)dy) = e^(-lny) = 1/y.
  4. Solve for x:
    x * (1/y) = ∫ 2ln(y) * (1/y) dy.
  5. Integrate and solve:
    Let u = ln(y), du = (1/y)dy. ∫2u du = u² + C = (ln y)² + C.
    So, x/y = (ln y)² + C, which gives x = y((ln y)² + C).

Example Problems