Homogeneous Differential Equations Advanced Quiz

1) The equation dy/dx = (x + y cos(y/x))/ (x cos(y/x)) is:

Linear

Homogeneous

Exact

Bernoulli

Let u = y/x, then the right side becomes (x + (u x) cos u)/(x cos u) = x (1 + u cos u)/(x cos u) = (1 + u cos u)/cos u, which depends only on u.

Explanation

Let u = y/x, then the right side becomes (x + (u x) cos u)/(x cos u) = x (1 + u cos u)/(x cos u) = (1 + u cos u)/cos u, which depends only on u.

2) Using u = y/x on dy/dx = (3x² + 2 x y + y²)/(x² + 3 x y), what is x du/dx?

(3 + 2u + u² - u(1 + 3u))/(1 + 3u)

(u² - 1)/(1 + 3u)

(2u² + 2u - 3)/(1 + 3u)

U + 2

Right side = (3x² + 2 x (u x) + (u x)²)/(x² + 3 x (u x)) = (3 + 2u + u²)/(1 + 3u). Then u + x du/dx = (3 + 2u + u²)/(1 + 3u), so x du/dx = (3 + 2u + u² - u - 3 u²)/(1 + 3u) = (3 + 2u + u² - u - 3u²)/(1 + 3u).

Explanation

Right side = (3x² + 2 x (u x) + (u x)²)/(x² + 3 x (u x)) = (3 + 2u + u²)/(1 + 3u). Then u + x du/dx = (3 + 2u + u²)/(1 + 3u), so x du/dx = (3 + 2u + u² - u - 3 u²)/(1 + 3u) = (3 + 2u + u² - u - 3u²)/(1 + 3u).

3) After substitution u = y/x in dy/dx = (4x + 3y)/(x + y), the equation becomes:

X du/dx = u + 3

Du/(u - 4) = dx/x

Du/(3u - 1) = dx/x

X du/dx = (3u + 4)/(u + 1) - u

dy/dx = (4 + 3u)/(1 + u), so u + x du/dx = (4 + 3u)/(1 + u), thus x du/dx = (4 + 3u)/(1 + u) - u.

Explanation

dy/dx = (4 + 3u)/(1 + u), so u + x du/dx = (4 + 3u)/(1 + u), thus x du/dx = (4 + 3u)/(1 + u) - u.

4) The isoclines of constant slope k for a homogeneous equation dy/dx = f(y/x) are

Straight lines through the origin

Circles

Parallel lines

Parabolas

Set f(y/x) = k, so y/x = constant (some function of k), so y = m x where m solves f(m) = k. Thus all isoclines are radial lines through the origin.

Explanation

Set f(y/x) = k, so y/x = constant (some function of k), so y = m x where m solves f(m) = k. Thus all isoclines are radial lines through the origin.

5) Which of the following is homogeneous of degree zero?

Dy/ddx = x² + y

Dy/dx = (y³ + x³)/(x y²)

Dy/dx = (x - y)/(x + y)

Dy/dx = y + 1

f(tx,ty) = (tx - ty)/(tx + ty) = t(x - y)/t(x + y) = (x - y)/(x + y) = f(x,y).

Explanation

f(tx,ty) = (tx - ty)/(tx + ty) = t(x - y)/t(x + y) = (x - y)/(x + y) = f(x,y).

6) The general solution to a homogeneous differential equation is y = x / (C - ln|x|). If the curve passes through the point (e, e), what is the value of C?

0

1

2

E

Substitute x = e and y = e into the solution. e = e / (C - ln|e|). Since ln|e| = 1, we have e = e / (C - 1). Dividing both sides by e gives 1 = 1 / (C - 1), so C - 1 = 1, which implies C = 2.

Explanation

Substitute x = e and y = e into the solution. e = e / (C - ln|e|). Since ln|e| = 1, we have e = e / (C - 1). Dividing both sides by e gives 1 = 1 / (C - 1), so C - 1 = 1, which implies C = 2.

7) After correct substitution, the solution to dy/dx = (x + y + 1)/(x - y - 2) contains:

Ln|x + y + 1|

A translation followed by homogeneous solution

Direct integration

Integrating factor

Let X = x - a, Y = y - b to eliminate constants. Solving gives the center at appropriate a,b, then solve homogeneous in new variables.

Explanation

Let X = x - a, Y = y - b to eliminate constants. Solving gives the center at appropriate a,b, then solve homogeneous in new variables.

8) The implicit solution to (2x + y) dx + (x + 2y) dy = 0 is:

X² + x y + y² = C

2x² + 2 x y + y² = C

X + y = C eˣ

Ln|2x + y| + ln|x + 2y| = K

dy/dx = -(2x + y)/(x + 2y). Let u = y/x, leads to du/(u² + u + 1) = dx/x, integration gives the quadratic form.

Explanation

dy/dx = -(2x + y)/(x + 2y). Let u = y/x, leads to du/(u² + u + 1) = dx/x, integration gives the quadratic form.

9) An advanced homogeneous equation is x² (y - x) dx + (x² + y²) dy = 0. The solution is:

Y² = C x

X y = C (x² + y²)

Ln|x| + arctan(y/x) = K

X² + y² = C x

Standard polar or homogeneous substitution leads to separable form with logarithmic and arctangent terms.

Explanation

Standard polar or homogeneous substitution leads to separable form with logarithmic and arctangent terms.

10) The differential equation dy/dx + y/x = (y/x)³ is:

Linear

Homogeneous

Quadratic

Exact

The differential equation dy/dx + y/x = (y/x)³ is classified as homogeneous. A first-order differential equation is homogeneous if it can be written in the form dy/dx = F(y/x), meaning the right-hand side depends only on the ratio y/x. Starting with the given equation, dy/dx + y/x = (y/x)³, we can rearrange it as dy/dx = (y/x)³ - y/x. The right-hand side is clearly a function of the combination y/x alone. To confirm, we use the standard substitution for homogeneous equations: let v = y/x, which implies y = vx. Differentiating both sides with respect to x gives dy/dx = v + x dv/dx. Substituting into the rearranged equation yields v + x dv/dx = v³ - v. Simplifying, we get x dv/dx = v³ - 2v. This resulting equation is separable and involves only v and x, which verifies that the original equation is homogeneous because it depends solely on the ratio v = y/x. The equation is not linear because the term (y/x)³ is nonlinear in y. And it is also not quadratic. It is not in the standard form of an exact equation without further manipulation.

Explanation

The differential equation dy/dx + y/x = (y/x)³ is classified as homogeneous. A first-order differential equation is homogeneous if it can be written in the form dy/dx = F(y/x), meaning the right-hand side depends only on the ratio y/x. Starting with the given equation, dy/dx + y/x = (y/x)³, we can rearrange it as dy/dx = (y/x)³ - y/x. The right-hand side is clearly a function of the combination y/x alone. To confirm, we use the standard substitution for homogeneous equations: let v = y/x, which implies y = vx. Differentiating both sides with respect to x gives dy/dx = v + x dv/dx. Substituting into the rearranged equation yields v + x dv/dx = v³ - v. Simplifying, we get x dv/dx = v³ - 2v. This resulting equation is separable and involves only v and x, which verifies that the original equation is homogeneous because it depends solely on the ratio v = y/x. The equation is not linear because the term (y/x)³ is nonlinear in y. And it is also not quadratic. It is not in the standard form of an exact equation without further manipulation.

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

X³ + y³ = C x y

X² + x y + y² = C

Ln|x/y| = C (x + y)

X/y = C

dy/dx = (1 - (y/x)³)/((y/x) + (y/x)²). Let u = y/x, du/(1 - u³) = dx/(x (u + u²)), correct separation and partial fractions give the cubic solution.

Explanation

dy/dx = (1 - (y/x)³)/((y/x) + (y/x)²). Let u = y/x, du/(1 - u³) = dx/(x (u + u²)), correct separation and partial fractions give the cubic solution.

12) The equation (y ln x - x ln y) dx + (x ln y - y ln x) dy = 0 is:

Exact

Homogeneous in ln terms

Linear

Separable

Let v = ln y / ln x = log_x y, or recognize it is homogeneous after considering the form.

Explanation

Let v = ln y / ln x = log_x y, or recognize it is homogeneous after considering the form.

13) The solution to dy/dx = (ax + by)/(cx + dy) where ad - bc ≠ 0 is generally:

Linear in x and y

Logarithmic or arctangent depending on discriminant

Always quadratic

Always transcendental

The sign of b² - 4ac in the characteristic determines whether the solution involves ln or arctan.

Explanation

The sign of b² - 4ac in the characteristic determines whether the solution involves ln or arctan.

14) For dy/dx = (2x² + 3xy + y²)/(x² + 2xy), the correct substitution yields:

A separable equation in u = y/x

Du/(u² + 3u + 2) = dx/x

Du/(u + 1)² = dx/x

Both A and C

Right side = (2 + 3u + u²)/(1 + 2u), then x du/dx = that minus u, simplifies to du/(u + 1)² = dx/x.

Explanation

Right side = (2 + 3u + u²)/(1 + 2u), then x du/dx = that minus u, simplifies to du/(u + 1)² = dx/x.

15) An advanced homogeneous equation of the form M(x,y) dx + N(x,y) dy = 0 where M and N are homogeneous of degree 3 has solution

Always F(x,y) = C where F is homogeneous of degree 4

Always logarithmic

Always involving arctan

Depends on the specific functions

For homogeneous equations of degree n, the solution is a homogeneous function of degree n+1 (or its logarithm if separation yields that form).

Explanation

For homogeneous equations of degree n, the solution is a homogeneous function of degree n+1 (or its logarithm if separation yields that form).

Homogeneous DEs: Advanced Forms, Nonlinear Variants & Techniques