Given
(3y−2xy3)dx+(4x−3x2y2)dy=0 then
M(x,y)=3y−2xy3N(x,y)=4x−3x2y2∂y∂M=3−6xy2∂x∂N=4−6xy2∂y∂M=∂x∂NShowing that the equation is not exact. Let
xayb be the integrating factor such that the given D.E will be exact. Multiplying the given equation through with the I.F gives;
(3xayb+1−2xa+1yb+3)dx+(4xa+1yb−3xa+2yb+2)dy
Since the equation is now exact, then
∂y∂M=∂x∂N where:
∂y∂M=3(b+1)xayb−2(b+3)xa+1yb+2∂x∂N=4(a+1)xayb−3(a+2)xa+1yb+2
By comparing the RHS of the equation, we get:
3(b+1)=4(a+1)⟹4a−3b=−1⋯eqn(i)−2(b+3)=−3(a+2)⟹b=23a⋯eqn(ii)
Solving the two equations simultaneously, we have:
a=2,b=3 Thus the IF for the equation to be exact is:
x2y3 Thus the exact form of the original equation is:
(3x2y4−2x3y5)dx+(4x3y3−3x4y5)dy=0 Integrating to get the solution to the equation:
∫(3x2y4−2x3y6)dx=x3y4−2x4y6∫(4x3y3−3x4y5)dy=x3y4−2x4y6 Therefore, the general solution of the DE is:
x3y4−2x4y6=c
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