Homogeneous Differential Equations: Basics, Substitution & Models

Setting v = y/x transforms dy/dx = f(y/x) into a separable equation because y = v x, dy/dx = v + x dv/dx, and the right side becomes a function of v only.

We can rewrite the equation for substitution as y = vx, on which we can use the product rule for differentiation to obtain dy/dx = v + x (dv/dx). Plugging this into the given equation gives v + x (dv/dx) = (x + vx)/(x - vx), which can be simplified into v + x (dv/dx) = (1 + v)/(1 - v). Subtract v from both sides to get x dv/dx = (1 + v)/(1 - v) - v, which is option b.

Replace x with tx and y with ty. f(tx, ty) = 3(tx)³ + 5(tx)²(ty) - (ty)³ = t³(3x³) + t³(5x²y) - t³(y³) = t³(3x³ + 5x²y - y³). Since the power of t is 3, the degree is 3.

Let v = y/x, so y = vx and dy/dx = v + x dv/dx. Substitute into the equation: v + x dv/dx = (vx - x)/(vx + x) = (v - 1)/(v + 1). Rearranging gives x dv/dx = (v - 1)/(v + 1) - v = (v - 1 - v(v + 1))/(v + 1) = (-1 - v²)/(v + 1) = -(1 + v²)/(v + 1). Separate variables: (v + 1)/(1 + v²) dv = -dx/x. Integrate both sides: ∫(v + 1)/(1 + v²) dv = -∫dx/x. The left side splits into ∫v/(1 + v²) dv + ∫1/(1 + v²) dv = (1/2) ln(1 + v²) + arctan(v). The right side is -ln|x| + constant. So we have (1/2) ln(1 + v²) + arctan(v) = -ln|x| + c. Multiply by 2: ln(1 + v²) + 2 arctan(v) = -2 ln|x| + 2c. Now substitute v = y/x: ln(1 + (y/x)²) + 2 arctan(y/x) = -2 ln|x| + 2c. But ln(1 + (y/x)²) = ln((x² + y²)/x²) = ln(x² + y²) - 2 ln|x|. Thus the equation becomes ln(x² + y²) - 2 ln|x| + 2 arctan(y/x) = -2 ln|x| + 2c. Cancel the -2 ln|x| terms to get ln(x² + y²) + 2 arctan(y/x) = 2c. Since 2c is an arbitrary constant, we can write the general solution as ln(x² + y²) + 2 arctan(y/x) = c.

m = 2x + y, n = -(x + 2y), both are degree 1 homogeneous functions, and f(tx,ty) = (2tx + ty)/(tx + 2ty) = same as original, so homogeneous.

v + x dv/dx = (4 - v)/(1 + v). Then x dv/dx = (4 - v)/(1 + v) - v = (4 - v - v - v²)/(1 + v) = (4 - 2v - v²)/(1 + v). This can be transformed into option c.

This is dy/dx = y/x, clearly homogeneous. Separate: dy/y = dx/x, integrate ln|y| = ln|x| + c, so y = c x.

The differential equation dy/dx = (x³ + y³)/(x y (x + y)) is homogeneous. To solve it, we use the substitution v = y/x, which means y = vx and dy/dx = v + x(dv/dx). First, let's rewrite the right-hand side by dividing both numerator and denominator by x³. The numerator becomes x³ + y³ = x³ + (vx)³ = x³(1 + v³), while the denominator becomes xy(x + y) = x(vx)(x + vx) = vx²(x(1 + v)) = vx³(1 + v). Therefore: dy/dx = (x³(1 + v³))/(vx³(1 + v)) = (1 + v³)/(v(1 + v)).

The lines x - 2y + 3 = 0 and 2x - y - 4 = 0 intersect at a point. Solving: let h = x - 4, k = y - 5 or similar, the intersection is found by solving the system, then set x = x - x0, y = y - y0 to make it homogeneous.

The given equation can be rearranged into dy/dx = (y² + xy)/x² = (y/x)² + (y/x). This is a homogeneous equation. Use the substitution v = y/x, so y = vx and dy/dx = v + x dv/dx. Substitute into the equation gives v + x dv/dx = v² + v. Which we simplify to obtain x dv/dx = v². Separating variables gives dv/v² = dx/x. Integrate both sides: ∫ dv/v² = ∫ dx/x, which gives -1/v = ln|x| + c1, where c1 is an arbitrary constant. Solve for v: v = -1/(ln|x| + c1). Substitute back v = y/x to get y/x = -1/(ln|x| + c1), so y = -x/(ln|x| + c1). Since c1 is an arbitrary constant, -c1 is also an arbitrary constant, which we can rename as c. This gives y = x / (c - ln|x|). For x > 0, we can write this as y = x/(c - ln x), which matches option c. The other options are not correct: option a, y³ + xy² = cx³, gives dy/dx = y/x, not (y/x)² + (y/x). Option b and d also do not satisfy the original differential equation.

dy/dx = c = y/x, so slope of orthogonal is -x/y. Solve dy/dx = -x/y, multiply both sides by y: y dy = -x dx, integrate (½)y² = - (½)x² + k, so x² + y² = constant.

The linear terms have constant parts, so the lines 3x + 4y + 1 = 0 and 4x + 3y + 2 = 0 intersect at one point. Solve the system 3x + 4y = -1, 4x + 3y = -2 to find the intersection, then shift origin there to make the new equation homogeneous.

Substitute v = y/x, so dy/dx = v + x dv/dx. The equation becomes v + x dv/dx = (3v²)/(2v - 1). Subtracting v from both sides gives x dv/dx = (3v²)/(2v - 1) - v = (3v² - 2v² + v)/(2v - 1) = (v² + v)/(2v - 1). Separating variables yields [(2v - 1)/(v² + v)] dv = dx/x. The left side requires partial fraction decomposition: (2v - 1)/(v(v + 1)) = A/v + B/(v + 1). Multiplying by v(v + 1) gives 2v - 1 = A(v + 1) + Bv. Setting v = 0 gives -1 = A(1), so A = -1. Setting v = -1 gives -3 = B(-1), so B = 3. Thus, P(v) = 3/(v + 1) - 1/v.

Let v = y/x, so that y = vx and dy/dx = v + x dv/dx. Substituting v = y/x into the given equation gives v + x dv/dx = (3v²) / (2v - 1). Then x dv/dx = (3v²) / (2v - 1) - v = (3v² - 2v² + v) / (2v - 1) = (v² + v) / (2v - 1). Separate variables to obtain (2v - 1) / (v(v + 1)) dv = dx/x. Using partial fraction decomposition on the left side: (2v - 1) / (v(v + 1)) = A/v + B/(v + 1). Solving for constants gives A = -1 and B = 3. Thus, the integral becomes ∫ (3/(v + 1) - 1/v) dv = ∫ dx/x. Integrating yields 3 ln|v + 1| - ln|v| = ln|x| + C. Combining logarithms gives ln| (v + 1)³ / v | = ln|x| + C. Exponentiating both sides: (v + 1)³ / v = K x. Substituting v = y/x back: (y/x + 1)³ / (y/x) = K x, which simplifies to ((x + y)/x)³ (x/y) = K x, and finally (x + y)³ = C x³ y.