Question:
Find the general solution of the differential equation \( (x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy \).
Find the general solution of the differential equation \( (x^3 - 3xy^2) dx = (y^3 - 3x^2y) dy \).
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Grouping terms to see if they form perfect total differentials can save you from lengthy calculations involved in substituting \(y = vx\) for homogeneous equations.
Updated On: Jul 8, 2026
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Solution and Explanation
Concept:
This equation can be identified as a homogeneous differential equation since all terms are of degree 3, or it can be solved by testing for an exact differential equation form \( M dx + N dy = 0 \).
• Exact Form condition: If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial y} \), the solution is given by \( \int M dx + \int (\text{terms of } N \text{ free from } x) dy = C \).
Step 1: Structuring the equation into differential form.
Rewrite the given equation as: \[ (x^3 - 3xy^2) dx - (y^3 - 3x^2y) dy = 0 \] Let \( M = x^3 - 3xy^2 \) and \( N = -(y^3 - 3x^2y) = 3x^2y - y^3 \).
Step 2: Checking for exactness.
Differentiate \( M \) partially with respect to \( y \): \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^3 - 3xy^2) = -6xy \] Differentiate \( N \) partially with respect to \( x \): \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3x^2y - y^3) = 6xy \] Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \) directly by sign, let's group the terms to integrate directly by grouping combinations: \[ x^3 dx - 3xy^2 dx + 3x^2y dy - y^3 dy = 0 \] Rearranging terms: \[ x^3 dx - y^3 dy - 3xy(y dx - x dy) = 0 \]
Step 3: Directly integrating individual components.
Multiply the whole differential equation by 4: \[ 4x^3 dx - 4y^3 dy - 12xy^2 dx + 12x^2y dy = 0 \] Notice that the total differential of the function \( f(x,y) = x^4 - 6x^2y^2 + y^4 \) is: \[ d(x^4 - 6x^2y^2 + y^4) = (4x^3 - 12xy^2)dx + (-12x^2y + 4y^3)dy \] Matching this with our grouped terms shows that integrating both sides directly yields: \[ x^4 - 6x^2y^2 + y^4 = C \]
• Exact Form condition: If \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial y} \), the solution is given by \( \int M dx + \int (\text{terms of } N \text{ free from } x) dy = C \).
Step 1: Structuring the equation into differential form.
Rewrite the given equation as: \[ (x^3 - 3xy^2) dx - (y^3 - 3x^2y) dy = 0 \] Let \( M = x^3 - 3xy^2 \) and \( N = -(y^3 - 3x^2y) = 3x^2y - y^3 \).
Step 2: Checking for exactness.
Differentiate \( M \) partially with respect to \( y \): \[ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^3 - 3xy^2) = -6xy \] Differentiate \( N \) partially with respect to \( x \): \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(3x^2y - y^3) = 6xy \] Since \( \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} \) directly by sign, let's group the terms to integrate directly by grouping combinations: \[ x^3 dx - 3xy^2 dx + 3x^2y dy - y^3 dy = 0 \] Rearranging terms: \[ x^3 dx - y^3 dy - 3xy(y dx - x dy) = 0 \]
Step 3: Directly integrating individual components.
Multiply the whole differential equation by 4: \[ 4x^3 dx - 4y^3 dy - 12xy^2 dx + 12x^2y dy = 0 \] Notice that the total differential of the function \( f(x,y) = x^4 - 6x^2y^2 + y^4 \) is: \[ d(x^4 - 6x^2y^2 + y^4) = (4x^3 - 12xy^2)dx + (-12x^2y + 4y^3)dy \] Matching this with our grouped terms shows that integrating both sides directly yields: \[ x^4 - 6x^2y^2 + y^4 = C \]
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