Homogeneous Differential Equations: Axis Shifts, Orthogonal Trajectories & Geometry

1) The differential equation dy/dx = (x – 3y)/(2x + y) is

Linear but not homogeneous

Homogeneous of degree 1

Bernoulli

Exact

Replace x → tx and y → ty: right-hand side becomes (tx – 3ty)/(2tx + ty) = t(x – 3y)/t(2x + y) = (x – 3y)/(2x + y), which is unchanged. The numerator and denominator are both degree-1 homogeneous polynomials, so the equation is homogeneous.

Explanation

Replace x → tx and y → ty: right-hand side becomes (tx – 3ty)/(2tx + ty) = t(x – 3y)/t(2x + y) = (x – 3y)/(2x + y), which is unchanged. The numerator and denominator are both degree-1 homogeneous polynomials, so the equation is homogeneous.

2) After setting v = y/x in dy/dx = (x + 4y)/(3x – y), the equation becomes

V + x dv/dx = (1 + 4v)/(3 – v)

V + x dv/dx = (1 – 4v)/(3 + v)

X dv/dx = (1 + 4v)/(3 + v) – v

X dv/dx = (4v – 1)/(3 – v)

Divide numerator and denominator by x: (x + 4y)/(3x – y) = (1 + 4(y/x))/(3 – y/x) = (1 + 4v)/(3 – v). Since v = y/x, dy/dx = v + x dv/dx, so the equation is exactly v + x dv/dx = (1 + 4v)/(3 – v).

Explanation

Divide numerator and denominator by x: (x + 4y)/(3x – y) = (1 + 4(y/x))/(3 – y/x) = (1 + 4v)/(3 – v). Since v = y/x, dy/dx = v + x dv/dx, so the equation is exactly v + x dv/dx = (1 + 4v)/(3 – v).

3) A differential equation of the form M(x, y)dx + N(x, y)dy = 0 is homogeneous if both M(x, y) and N(x, y) are homogeneous functions of the same degree.

True

False

This is the definition of a homogeneous equation in differential form. If M and N are both homogeneous of degree n, then the function f(x,y) = -M(x,y)/N(x,y) will be homogeneous of degree zero, meaning f(tx, ty) = f(x,y). This satisfies the condition for the equation dy/dx = f(x,y) to be solved using the substitution v = y/x.

Explanation

This is the definition of a homogeneous equation in differential form. If M and N are both homogeneous of degree n, then the function f(x,y) = -M(x,y)/N(x,y) will be homogeneous of degree zero, meaning f(tx, ty) = f(x,y). This satisfies the condition for the equation dy/dx = f(x,y) to be solved using the substitution v = y/x.

4) To solve the equation dy/dx = (x + y + 1) / (x + y - 2), why can you NOT use the standard translation of axes (X = x - h, Y = y - k)?

Because the lines x + y + 1 = 0 and x + y - 2 = 0 are parallel.

Because the constant terms are not zero.

Because the coefficients of x are equal.

Because the equation is already homogeneous.

Parallel lines do not intersect, so there is no point (h,k) to shift the origin to.To find an intersection point to shift the origin to, we would solve the system x + y + 1 = 0 and x + y - 2 = 0. These lines have the same coefficients (1,1) for x and y, so they are parallel. Since parallel lines do not intersect, there is no point (h,k) to shift the origin to.

Explanation

Parallel lines do not intersect, so there is no point (h,k) to shift the origin to.To find an intersection point to shift the origin to, we would solve the system x + y + 1 = 0 and x + y - 2 = 0. These lines have the same coefficients (1,1) for x and y, so they are parallel. Since parallel lines do not intersect, there is no point (h,k) to shift the origin to.

5) The isoclines for any homogeneous differential equation dy/dx = f(y/x) are:

Concentric circles centered at the origin.

Straight lines passing through the origin.

Hyperbolas xy = C.

Parallel lines with slope 1.

An isocline is defined by dy/dx = k. For a homogeneous equation, this means f(y/x) = k. Assuming f is invertible, y/x = c (a constant), which implies y = cx. The graph of y = cx is a straight line through the origin.

Explanation

An isocline is defined by dy/dx = k. For a homogeneous equation, this means f(y/x) = k. Assuming f is invertible, y/x = c (a constant), which implies y = cx. The graph of y = cx is a straight line through the origin.

6) The equation dy/dx = (3x² + 2xy)/(x² + 2xy + y²) in terms of v = y/x is

(3 + 2v)/(1 + 2v + v²)

(3v + 2)/(v² + 2v + 1)

V(3 + 2v)/(1 + 2v + v²)

3 + 2v

Divide numerator and denominator by x²: numerator (3x² + 2xy)/x² = 3 + 2(y/x) = 3 + 2v, denominator (x² + 2xy + y²)/x² = 1 + 2(y/x) + (y/x)² = 1 + 2v + v². The equation is clearly homogeneous of degree 2, and the form is exactly (3 + 2v)/(1 + 2v + v²).

Explanation

Divide numerator and denominator by x²: numerator (3x² + 2xy)/x² = 3 + 2(y/x) = 3 + 2v, denominator (x² + 2xy + y²)/x² = 1 + 2(y/x) + (y/x)² = 1 + 2v + v². The equation is clearly homogeneous of degree 2, and the form is exactly (3 + 2v)/(1 + 2v + v²).

7) Find the particular solution to (x² + y²) dx + 2xy dy = 0 subject to y(1) = -1.

X(y² + 3x²) = 4

3x y² + x³ = 4

3x y² + x³ = 2

Y² = -x²

This is a homogeneous equation. Using substitution y=vx, it simplifies to the integrable form leading to 3xy² + x³ = C. To find C, substitute the point (1, -1): 3(1)(-1)² + (1)³ = 3(1)(1) + 1 = 3 + 1 = 4. Therefore, C = 4, and the particular solution is 3xy² + x³ = 4.

Explanation

This is a homogeneous equation. Using substitution y=vx, it simplifies to the integrable form leading to 3xy² + x³ = C. To find C, substitute the point (1, -1): 3(1)(-1)² + (1)³ = 3(1)(1) + 1 = 3 + 1 = 4. Therefore, C = 4, and the particular solution is 3xy² + x³ = 4.

8) The orthogonal trajectories to the family xy = C are

X² + y² = K

X² – y² = K

(x + y)² = K

X/y = constant

From xy = C, differentiate implicitly: x dy/dx + y = 0, so dy/dx = –y/x. Orthogonal trajectories have slope x/y. Solve dy/dx = x/y → y dy = x dx → (½)y² = (½)x² + constant → x² + y² = K (circles centered at origin).

Explanation

From xy = C, differentiate implicitly: x dy/dx + y = 0, so dy/dx = –y/x. Orthogonal trajectories have slope x/y. Solve dy/dx = x/y → y dy = x dx → (½)y² = (½)x² + constant → x² + y² = K (circles centered at origin).

9) The equation dy/dx = (2x + 3y – 5)/(3x – 4y + 2) is

Already homogeneous

Becomes homogeneous after translation of variables

Linear

Bernoulli

The arguments 2x + 3y – 5 and 3x – 4y + 2 are linear with constant terms. The lines 2x + 3y = 5 and 3x – 4y = –2 intersect at one point (solve simultaneously to get x = 1, y = 1). Setting X = x – 1, Y = y – 1 removes the constants and makes the equation homogeneous in X and Y.

Explanation

The arguments 2x + 3y – 5 and 3x – 4y + 2 are linear with constant terms. The lines 2x + 3y = 5 and 3x – 4y = –2 intersect at one point (solve simultaneously to get x = 1, y = 1). Setting X = x – 1, Y = y – 1 removes the constants and makes the equation homogeneous in X and Y.

10) After applying the translation x = X + 1, y = Y - 1 to the equation dy/dx = (x + y) / (x - y - 2), we obtain the homogeneous equation dY/dX = (X + Y) / (X - Y). What is the relationship between the solution curve in the XY-plane and the solution in the xy-plane?

Shifted 1 right and 1 down

Shifted 1 left and 1 up

Rotated 45°

Scaled by 2

Since x = X + 1, the "new" coordinate x is the "old" coordinate X plus 1. This is a shift of the origin. If the center of the solution in XY is (0,0), the center in xy is (1, -1). The vector (1, -1) represents a shift Right 1, Down 1.

Explanation

Since x = X + 1, the "new" coordinate x is the "old" coordinate X plus 1. This is a shift of the origin. If the center of the solution in XY is (0,0), the center in xy is (1, -1). The vector (1, -1) represents a shift Right 1, Down 1.

11) The general solution of dy/dx = (x³ – y³)/(x² y + x y²) is

X³ + y³ = C x y

X – y = C x² y²

X² + x y + y² = C

Ln|x/y| = C x y

Simplify first: dy/dx = (x³ – y³)/(x y (x + y)) = (x² – x y + y²)/(x y). Better in v: divide by x³, get (1 – v³)/(v + v²) = (1 – v³)/[v(1 + v)]. Then v + x dv/dx = (1 – v³)/[v(1 + v)]. Standard solution procedure yields x³ + y³ = C x y.

Explanation

Simplify first: dy/dx = (x³ – y³)/(x y (x + y)) = (x² – x y + y²)/(x y). Better in v: divide by x³, get (1 – v³)/(v + v²) = (1 – v³)/[v(1 + v)]. Then v + x dv/dx = (1 – v³)/[v(1 + v)]. Standard solution procedure yields x³ + y³ = C x y.

12) Solve the homogeneous differential equation: (x² + y²) dx - 2xy dy = 0.

X² + y² = C

X² - y² = C x

X² + y² = C x

X² - y² = C

The given equation is homogeneous because both terms are polynomials of degree 2. To solve, use the substitution y = vx, so dy = v dx + x dv. Substitute into the equation: (x² + v² x²) dx - 2x(vx)(v dx + x dv) = 0. Simplify by factoring out x²: x²(1+v²) dx - 2x² v (v dx + x dv) = 0. Divide by x² (assuming x ≠ 0): (1+v²) dx - 2v(v dx + x dv) = 0. Expand: (1+v²) dx - 2v² dx - 2v x dv = 0. Combine like terms: (1 - v²) dx - 2v x dv = 0. Separate variables: (1 - v²) dx = 2v x dv which implies dx/x = [2v/(1-v²)] dv. Integrate both sides: ∫ dx/x = ∫ [2v/(1-v²)] dv. The left side gives ln|x|. The right side integrates to -ln|1-v²|, because the derivative of 1-v² is -2v. So, ln|x| = -ln|1-v²| + constant. Combine logarithms: ln|x| + ln|1-v²| = ln|C|, where C is a constant. This gives ln|x(1-v²)| = ln|C|, so x(1-v²) = C. Substitute back v = y/x: x(1 - (y/x)²) = C to get x(1 - y²/x²) = C. This leads to x - y²/x = C. Multiply through by x to get x² - y² = C x.

Explanation

The given equation is homogeneous because both terms are polynomials of degree 2. To solve, use the substitution y = vx, so dy = v dx + x dv. Substitute into the equation: (x² + v² x²) dx - 2x(vx)(v dx + x dv) = 0. Simplify by factoring out x²: x²(1+v²) dx - 2x² v (v dx + x dv) = 0. Divide by x² (assuming x ≠ 0): (1+v²) dx - 2v(v dx + x dv) = 0. Expand: (1+v²) dx - 2v² dx - 2v x dv = 0. Combine like terms: (1 - v²) dx - 2v x dv = 0. Separate variables: (1 - v²) dx = 2v x dv which implies dx/x = [2v/(1-v²)] dv. Integrate both sides: ∫ dx/x = ∫ [2v/(1-v²)] dv. The left side gives ln|x|. The right side integrates to -ln|1-v²|, because the derivative of 1-v² is -2v. So, ln|x| = -ln|1-v²| + constant. Combine logarithms: ln|x| + ln|1-v²| = ln|C|, where C is a constant. This gives ln|x(1-v²)| = ln|C|, so x(1-v²) = C. Substitute back v = y/x: x(1 - (y/x)²) = C to get x(1 - y²/x²) = C. This leads to x - y²/x = C. Multiply through by x to get x² - y² = C x.

13) Isoclines for the homogeneous equation dy/dx = (ax + by)/(cx + dy) are

Ax + by = k

Circles

Lines through origin

Parallel lines

Set (ax + by)/(cx + dy) = k, then ax + by = k(cx + dy), (a – k c)x + (b – k d)y = 0, which is a family of straight lines passing through the origin (for varying k).

Explanation

Set (ax + by)/(cx + dy) = k, then ax + by = k(cx + dy), (a – k c)x + (b – k d)y = 0, which is a family of straight lines passing through the origin (for varying k).

14) The differential equation (x² + 3xy + 2y²) dx + (x² + xy) dy = 0 has the integrating factor

1/(xy)

1/(x²y)

1/(x+y)²

None

Although the equation is homogeneous (degree 3/degree 3), checking exactness and standard integrating factors of the form 1/(x y) or similar fails. The fastest method is v = y/x, leading to a separable equation.

Explanation

Although the equation is homogeneous (degree 3/degree 3), checking exactness and standard integrating factors of the form 1/(x y) or similar fails. The fastest method is v = y/x, leading to a separable equation.

15) Advanced: The curves orthogonal to all circles passing through the origin are

Straight lines through origin

Circles with diameter on x-axis

Another family of circles

Y = k x²

Circles through the origin have equation x² + y² + d x + e y = 0. Differentiate implicitly: 2x + 2y dy/dx + d + e dy/dx = 0, so dy/dx = –(2x + d)/(2y + e). The orthogonal slope is (2y + e)/(2x + d). Solving this homogeneous equation gives radial straight lines y = k x (lines through the origin).

Explanation

Circles through the origin have equation x² + y² + d x + e y = 0. Differentiate implicitly: 2x + 2y dy/dx + d + e dy/dx = 0, so dy/dx = –(2x + d)/(2y + e). The orthogonal slope is (2y + e)/(2x + d). Solving this homogeneous equation gives radial straight lines y = k x (lines through the origin).