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Question 101 of 126

Q.Solve: (x3+3xy2)dx+(y3+3x2y)dy=0(x^3 + 3xy^2)dx + (y^3 + 3x^2y)dy = 0

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★
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Checking exactness (My=Nx=6xyM_y=N_x=6xy) and integrating MM w.r.t. xx (fixing the function of yy via matching against NN) gives the implicit solution x4+6x2y2+y4=Cx^4+6x^2y^2+y^4=C.

  1. Write the equation as M dx+N dy=0M\,dx+N\,dy=0 with M=x3+3xy2M=x^3+3xy^2 and N=y3+3x2yN=y^3+3x^2y.
  2. Test for exactness: ∂M∂y=∂∂y(x3+3xy2)=6xy\dfrac{\partial M}{\partial y} = \dfrac{\partial}{\partial y}(x^3+3xy^2) = 6xy.
  3. ∂N∂x=∂∂x(y3+3x2y)=6xy\dfrac{\partial N}{\partial x} = \dfrac{\partial}{\partial x}(y^3+3x^2y) = 6xy.
  4. Since ∂M∂y=∂N∂x=6xy\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}=6xy, the equation is exact, so a function F(x,y)F(x,y) exists with Fx=MF_x=M, Fy=NF_y=N and general solution F(x,y)=CF(x,y)=C.
  5. Integrate MM with respect to xx, holding yy constant: F=∫(x3+3xy2) dx=x44+3x2y22+g(y)F = \displaystyle\int (x^3+3xy^2)\,dx = \dfrac{x^4}{4}+\dfrac{3x^2y^2}{2} + g(y), where g(y)g(y) is an arbitrary function of yy (the 'constant' of this partial integration).
  6. Differentiate this FF with respect to yy: ∂F∂y=3x2y+g′(y)\dfrac{\partial F}{\partial y} = 3x^2y + g'(y). …

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