Given that the non-exact DE
$\left(5 x y^2-2 y\right) d x+\left(3 x^2 y-x\right) d y=0$
has an $I F$ in the form $\mu(x, y)=x^a y^b$ for some positive integers $a$ and $b$, find the general solution of the equation.

Find the integrating factor μ ( x , y ) = x a y b by equating the partial derivatives after multiplying the original equation by μ ( x , y ) , which gives a = 3 and b = 1 , so μ ( x , y ) = x 3 y .
Multiply the original equation by the integrating factor to obtain the exact equation ( 5 x 4 y 3 − 2 x 3 y 2 ) d x + ( 3 x 5 y 2 − x 4 y ) d y = 0 .
Integrate M ( x , y ) with respect to x to find F ( x , y ) = x 5 y 3 − 2 1 ​ x 4 y 2 + g ( y ) , and then find g ( y ) by comparing ∂ y ∂ F ​ with N ( x , y ) .
The general solution is x 5 y 3 − 2 1 ​ x 4 y 2 = C .

x 5 y 3 − 2 1 ​ x 4 y 2 = C ​
Explanation

Problem Setup We are given the differential equation ( 5 x y 2 − 2 y ) d x + ( 3 x 2 y − x ) d y = 0 and an integrating factor of the form μ ( x , y ) = x a y b . Our goal is to find the general solution of the equation.

Multiply by Integrating Factor First, we multiply the differential equation by the integrating factor μ ( x , y ) = x a y b :
( 5 x y 2 − 2 y ) x a y b d x + ( 3 x 2 y − x ) x a y b d y = 0 ( 5 x a + 1 y b + 2 − 2 x a y b + 1 ) d x + ( 3 x a + 2 y b + 1 − x a + 1 y b ) d y = 0

Compute Partial Derivatives For the equation to be exact, we need to satisfy the condition ∂ y ∂ M ​ = ∂ x ∂ N ​ , where M ( x , y ) = 5 x a + 1 y b + 2 − 2 x a y b + 1 and N ( x , y ) = 3 x a + 2 y b + 1 − x a + 1 y b . Let's compute the partial derivatives: ∂ y ∂ M ​ = 5 ( b + 2 ) x a + 1 y b + 1 − 2 ( b + 1 ) x a y b ∂ x ∂ N ​ = 3 ( a + 2 ) x a + 1 y b + 1 − ( a + 1 ) x a y b

Equate Partial Derivatives Now, we equate the partial derivatives: 5 ( b + 2 ) x a + 1 y b + 1 − 2 ( b + 1 ) x a y b = 3 ( a + 2 ) x a + 1 y b + 1 − ( a + 1 ) x a y b

Solve for a and b To find a and b , we equate the coefficients of the terms x a + 1 y b + 1 and x a y b :
5 ( b + 2 ) = 3 ( a + 2 ) and 2 ( b + 1 ) = ( a + 1 ) From the second equation, we have a = 2 b + 1 . Substituting this into the first equation: 5 ( b + 2 ) = 3 ( 2 b + 1 + 2 ) 5 b + 10 = 6 b + 9 b = 1 Then, a = 2 ( 1 ) + 1 = 3 . So, a = 3 and b = 1 .

Find Integrating Factor The integrating factor is μ ( x , y ) = x 3 y 1 = x 3 y . Multiplying the original differential equation by this integrating factor, we get: ( 5 x y 2 − 2 y ) x 3 y d x + ( 3 x 2 y − x ) x 3 y d y = 0 ( 5 x 4 y 3 − 2 x 3 y 2 ) d x + ( 3 x 5 y 2 − x 4 y ) d y = 0

Solve Exact Equation Now, we have M ( x , y ) = 5 x 4 y 3 − 2 x 3 y 2 and N ( x , y ) = 3 x 5 y 2 − x 4 y . We want to find a function F ( x , y ) such that ∂ x ∂ F ​ = M ( x , y ) and ∂ y ∂ F ​ = N ( x , y ) .
Integrating M ( x , y ) with respect to x , we get: F ( x , y ) = ∫ ( 5 x 4 y 3 − 2 x 3 y 2 ) d x = x 5 y 3 − 2 1 ​ x 4 y 2 + g ( y ) Now, we differentiate F ( x , y ) with respect to y :
∂ y ∂ F ​ = 3 x 5 y 2 − x 4 y + g ′ ( y ) Comparing this with N ( x , y ) = 3 x 5 y 2 − x 4 y , we see that g ′ ( y ) = 0 , so g ( y ) = C 1 ​ , where C 1 ​ is a constant.

General Solution Thus, the general solution is given by: F ( x , y ) = x 5 y 3 − 2 1 ​ x 4 y 2 = C where C is an arbitrary constant.


Examples
In fluid dynamics, understanding and solving differential equations is crucial for modeling fluid flow. For instance, when analyzing the flow of a viscous fluid in a pipe, the governing equations often involve non-exact differential equations. By finding an appropriate integrating factor, engineers can transform these equations into exact forms, making them solvable and allowing for the prediction of fluid behavior under various conditions. This is essential for designing efficient pipelines and optimizing fluid transport processes.

Answered by GinnyAnswer | 2025-07-05