University:
Kentucky State University
Course:
MAT 115 | College Algebra
Academic year:

2024
Views:

338

Pages:

8
Author:

kleuk3j4

**2.2: Homogeneous Equations** Def: The equation dy/dx = f(x, y) is said to be homogeneous if f(x, y) can be expressed as a function of (y/x). **Example 1:** dy/dx = (x+y)/(x-y) => dy/dx = ((x+y)/x)/((x-y)/x) = (1+(y/x))/(1-(y/x)) so it is homogeneous. **Example 2:** dy/dx = x/(x+y) => dy/dx = 1/(1+(y/x)) => hom. **Example 3:** dy/dx = (x²+xy+y²)/x² => dy/dx = 1+(y/x)+(y/x)² => hom. **Example 4:** xy' + y = x²cos(y/x) => y' + (y/x) = cos(y/x) => hom. **Example 5:** y' = (x+y)/(x-y) + ln(y/x) => y' = (x/x + y/x)/(x/x - y/x) + ln(y/x) => hom. **Example 6:** y' = (x+3y)/(2x+3y) y' = 1/(2+3(y/x)), non-hom. If the equation dy/dx = f(x,y) is homogeneous, that is, dy/dx = g(y/x) then we can transform it to a separable equation and solve it. Let y/x = v => y = xv dy = x dv + v dx or dy/dx = v + x dv/dx => v + x dv/dx = g(v) x dv/dx = g(v) - v dv/[g(v)-v] = dx/x separable. **Example 7:** Solve xy' = y + x cos²(y/x) y' = y/x + cos²(y/x) - homogeneous Let y/x = v => dy = x dv + v dx => v + x dv/dx = v + cos²(v) x dv/dx = cos²(v) ∫sec²v dv = ∫dx/x tan v = ln|x| + c v = tan⁻¹[ln|x| + c] v = y/x y = x tan⁻¹[ln|x| + c] general solution. The general solution: y = x - x / sin[ln|x| + c] **Example 10:** Solve xy' = y + √(x²-y²) for x > 0 y' = (y/x) + √(1-(y/x)²) Let y/x = v => dy = v dx + x dv => v + x dv/dx = v + √(1-v²) x dv/dx = √(1-v²) ∫ 1/√(1-v²) dv = ∫ dx/x => sin⁻¹v = ln|x| + c v = sin[ln|x| + c] v = y/x y = x sin[ln|x| + c] **Example 11:** Transform xy' = 2y-x+5 into homogeneous. Let u = x+a, y = v+b dy/dx = [2(v+b)-(u-a)+5] / [2u-v+b-4] => a-2b+5 = 0 } => -2a+b=4 -2a+b-4 = 0 } => 2a-4b=10 b = 2 and a = -1 dy = dv and dx = du dy/dx = dv/du = [2v-u] / [2u-v] is homogeneous

Homogeneous Equations

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