Solving First Order and First Degree Differential Equations

Solving First Order and First Degree Differential Equations

Author: Bindeshwar Singh Kushwaha
Institute: PostNetwork Academy

Reducing to Homogeneous Form

• Consider the differential equation of the form:
  $$\frac{dy}{dx} = \frac{a x + b y + c}{a’ x + b’ y + c’}$$
• To reduce it to homogeneous form, use the substitution:
  $$x = X + h, \quad y = Y + k$$
• Choose constants \( h \) and \( k \) such that the equation becomes homogeneous in \( X \) and \( Y \).
• Substituting yields:
  $$\frac{dY}{dX} = \frac{aX + bY + (ah + bk + c)}{a’X + b’Y + (a’h + b’k + c’)}$$
• To make it homogeneous:
  $$ah + bk + c = 0, \quad a’h + b’k + c’ = 0$$
• Solve these simultaneous equations for \( h \) and \( k \).

Special Cases and Limitations

• If:
  $$ah + bk + c = 0, \quad a’h + b’k + c’ = 0$$
  has no unique solution, then:
• For a unique solution, the determinant must be non-zero:
  $$\frac{a}{a’} \ne \frac{b}{b’} \quad \text{or} \quad ab’ \ne a’b$$
• If all terms are proportional:
  $$\frac{a}{a’} = \frac{b}{b’} = \frac{c}{c’}$$
  then:
  $$\frac{dy}{dx} = 1 \Rightarrow y = x + C$$
• If no simplification works, try integrating factors or symmetry-based substitutions.

Example 1: Solve \( \frac{dy}{dx} = \frac{x – 4y – 1}{x + y + 5} \)

• Given:
  $$\frac{dy}{dx} = \frac{x – 4y – 1}{x + y + 5}$$
• Make substitution:
  $$x = X – 2, \quad y = Y – 3$$
• Then:
  $$\frac{dY}{dX} = \frac{X – 4Y}{X + Y}$$
• Let \( Y = vX \) then:
  $$v + X \frac{dv}{dX} = \frac{1 – 4v}{1 + v}$$
• Simplifying:
  $$\frac{dv}{dX} = \frac{1 – 5v – v^2}{X(1 + v)}$$
• Separate and integrate:
  $$\int \frac{1 + v}{1 – 5v – v^2} \, dv = \int \frac{1}{X} \, dX$$
• Right-hand side:
  $$\ln |X| + c$$

Final Answer of Example 1

• Left-hand side:
  $$\frac{1}{2} \ln(1 – 5v – v^2) – \tan^{-1}(v + 2.5) + c$$
• Back-substitute \( v = \frac{Y}{X} \), \( Y = y + 3 \), \( X = x + 2 \)
• Final answer:
  $$\frac{1}{2} \ln\left[(x + 2)^2 + (y + 3)^2\right] + \tan^{-1}\left(\frac{y + 3}{x + 2}\right) = c$$

Example 2: Solve \( (4x + 6y + 5)\,dy = (3y + 2x + 5)\,dx \)

• Rewrite:
  $$\frac{dy}{dx} = \frac{3y + 2x + 5}{4x + 6y + 5}$$
• Let \( v = 2x + 3y \), then:
  $$\frac{dy}{dx} = \frac{v + 5}{2v + 5}$$
• So:
  $$\frac{dv}{dx} = \frac{2(v + 5)}{2v + 5}$$
• Separate:
  $$\frac{2v + 5}{v + 5} \, dv = dx$$
• Algebra:
  $$\frac{2v + 5}{v + 5} = 2 – \frac{5}{v + 5}$$
• Integrate:
  $$2v – 5 \ln |v + 5| = x + c$$
• Back-substitute:
  $$2(2x + 3y) – 5 \ln |2x + 3y + 5| = x + c$$

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Equations Reducible to Homogeneous Form

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