Question

differential equation (3x^2y^4+2xy)dx+(2x^3y^3-x^2)dy=0

Answer 1

The correct question is : Show that following differential equation is not exact.

(3x^2y^4+2xy)dx+(2x^3y^3-x^2)dy=0

Then find an integrating factor to solve the differential equation.

Answer:

Given: the given equation is (3x^2y^4 + 2xy)dx + (2x^3y^3 - x^2)dy = 0

To find: Now we have to find the integrating factor of above equation.

Solution:

(3x^2y^4+2xy)dx+(2x^3y^3-x^2)dy=0  .............(1)

let, M = 3x^2 * y^4+2xy

    dM/dy =  12x²y³ + 2x

and N = 2x^3 * y^3-x^2

      dN/dx = 6x²y³ - 2x

Since dM/dy  ≠ dN/dx , therefore DE is not exact .

ù(y) = exp ∫ [ (dN/dx - dM/dy ) / M ] dy

ù(y) = exp ∫ [ ( 6x²y³ - 2x - 12x²y³ - 2x ) / (3x²y⁴ +2xy ) ] dy

ù(y) = exp ∫ [ (-6x²y³ - 4x) / (3x²y⁴ +2xy ) ] dy

ù(y) = exp ∫ -2/y dy = 1/y²

multiply equation 1 on both side by ù(y) = 1/y² , we get

( 3x²y² + 2x/y ) dx  + ( 2x³y - x²/y² ) dy = 0

dM/dy = 6x²y -2x/y²                    

dN/dx = 6x²y -2x/y²

dM/dy = dN/dx , DE is exact...

df/dx = 3x²y² + 2x/y  ..................(i)

df/dy = 2x³y - x²/y²  ................(ii)

integrate (i) wrt x , we get

f(x,y) = x³y³ + x²/y + h(y)  ................(iii)

differentiate above wrt y , we get

df/dy = 2x³y - x²/y² + h'(y)

compare above and (ii) , we get

h'(y) = 0

integrate both sides wrt y, we get h(y) = c₁

therefore, eq (iii) becomes

f(x,y) = x³y² + x²/y = c1 = 0

          x³y² + x²/y = c

note: In mathematics, a differential equation is an equation that creates a relationship between one or more functions and their derivatives. It plays an important role in many fields that are engineering, physics, economics, and biology, etc.