Question:

The general solution of the differential equation: (6x22xy18x+3y)dx(x23x)dy=0 (6x^2 - 2xy - 18x + 3y) dx - (x^2 - 3x) dy = 0  

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For exact differential equations, check if My=Nx \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} , then integrate M(x,y) M(x, y) with respect to x x , match terms using N(x,y) N(x, y) , and solve for the general solution.
Updated On: Mar 17, 2025
  • 2x3x2y9x2+3y+C=0 2x^3 - x^2 y - 9x^2 + 3y + C = 0
  • 4x32x2y6x2+6xy+C=0 4x^3 - 2x^2 y - 6x^2 + 6xy + C = 0
  • 2x24xyy2x+3y+C=0 2x^2 - 4xy - y^2 - x + 3y + C = 0
  • 3x2+5xy2y24x2y+C=0 3x^2 + 5xy - 2y^2 - 4x - 2y + C = 0 ]

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The Correct Option is A

Solution and Explanation

Step 1: Checking for Exactness Given: M(x,y)=6x22xy18x+3y,N(x,y)=(x23x) M(x, y) = 6x^2 - 2xy - 18x + 3y, \quad N(x, y) = -(x^2 - 3x) Compute: My=2x+3,Nx=2x+3 \frac{\partial M}{\partial y} = -2x + 3, \quad \frac{\partial N}{\partial x} = -2x + 3 Since My=Nx \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} , the equation is exact. 

Step 2: Solving for F(x,y) F(x, y) Integrate M(x,y) M(x, y) with respect to x x : F(x,y)=(6x22xy18x+3y)dx F(x, y) = \int (6x^2 - 2xy - 18x + 3y) dx =2x3x2y9x2+3yx+g(y) = 2x^3 - x^2 y - 9x^2 + 3yx + g(y) Differentiate with respect to y y : dFdy=x2+3x+g(y) \frac{dF}{dy} = -x^2 + 3x + g'(y) Since dFdy=N(x,y) \frac{dF}{dy} = N(x, y) , equating: x2+3x+g(y)=x2+3x -x^2 + 3x + g'(y) = -x^2 + 3x g(y)=0g(y)=C g'(y) = 0 \Rightarrow g(y) = C  

Step 3: Final Equation 2x3x2y9x2+3yx+C=0 2x^3 - x^2 y - 9x^2 + 3yx + C = 0  

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