Solving First Order and First Degree Differential Equations
Author: Bindeshwar Singh Kushwaha
Institute: PostNetwork Academy
Example 1: Solve \( \frac{dy}{dx} = e^x \)
- Equation: \( \frac{dy}{dx} = e^x \)
- Multiply both sides by \( dx \): \( dy = e^x dx \)
- Integrate: \( \int dy = \int e^x dx \)
- Solution: \( y = e^x + c \)
- General solution: \( y(x) = e^x + c \)
Example 2: Initial Value Problem
- Given: \( (1 + y^2)dx + (1 + x^2)dy = 0,\quad y(0) = -1 \)
- Rewrite: \( \frac{dx}{1 + x^2} + \frac{dy}{1 + y^2} = 0 \)
- Integrate: \( \tan^{-1} x + \tan^{-1} y = c \)
- Using initial condition: \( \tan^{-1}(0) + \tan^{-1}(-1) = -\frac{\pi}{4} \)
- Solution: \( \tan^{-1} x + \tan^{-1} y = -\frac{\pi}{4} \)
Example 3: Solve Initial Value Problem
- \( \frac{dy}{dx} = \frac{3x^2 + 4x + 2}{2(y – 1)} \), \( y \ne 1 \), \( y(0) = -1 \)
- Rewriting: \( 2(y – 1)dy = (3x^2 + 4x + 2)dx \)
- Integration: \( y^2 – 2y = x^3 + 2x^2 + 2x + c \)
- Using initial values: \( 1 + 2 = 3 \Rightarrow c = 3 \)
- Solution: \( y^2 – 2y = x^3 + 2x^2 + 2x + 3 \)
- Solving for \( y \): \( y = 1 \pm \sqrt{x^3 + 2x^2 + 2x + 4} \)
Choosing Correct Root
- Use \( y(0) = -1 \): \( y = 1 – \sqrt{4} = -1 \)
- Final solution: \( y = 1 – \sqrt{x^3 + 2x^2 + 2x + 4} \)
Domain of Validity
- Ensure \( \sqrt{x^3 + 2x^2 + 2x + 4} \) is real
- Valid for \( x > -2 \)
Definition: Homogeneous Function
- A function \( h(x, y) \) is homogeneous of degree \( n \) if:
- \( h(\lambda x, \lambda y) = \lambda^n h(x, y) \) for all \( \lambda > 0 \)
Question: Solve \( \frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \), \( x > 0 \)
- Homogeneous equation: \( \frac{dy}{dx} = 2 \left( \frac{y}{x} \right)^2 + 3 \left( \frac{y}{x} \right) \)
- Use substitution: \( v = \frac{y}{x},\; y = vx,\; \frac{dy}{dx} = v + x \frac{dv}{dx} \)
- Results in: \( \frac{dv}{dx} = \frac{v(2v + 1)}{x} \)
- Separate: \( \frac{dv}{v(v+1)} = \frac{2dx}{x} \)
- Partial fractions: \( \frac{1}{v(v+1)} = \frac{1}{v} – \frac{1}{v+1} \)
- Integrate: \( \ln\left|\frac{v}{v+1}\right| = \ln|cx^2| \)
- Back-substitute: \( \frac{y}{x+y} = cx^2 \Rightarrow y = \frac{cx^3}{1 – cx^2} \)
Question: Solve \( \frac{dy}{dx} = \frac{y^3}{x^3} + \frac{y}{x} \), \( x > 0 \)
- Homogeneous equation, use \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \)
- Simplifies to: \( x \frac{dv}{dx} = v^3 \)
- Separate: \( \int \frac{1}{v^3} dv = \int \frac{1}{x} dx \)
- Integration: \( -\frac{1}{2v^2} = \ln x + \ln |c| \)
- Back-substitute: \( v = \frac{y}{x} \Rightarrow y^2 = -\frac{x^2}{2 \ln(x|c|)} \)