(4ye^3x - 6)dx + (e^3x (4x + 6y^5))dy = 0 in differential form M^~ dx + N^~ dy =

Question

(4ye^3x - 6)dx + (e^3x (4x + 6y^5))dy = 0 in differential form M^~ dx + N^~ dy = 0 is not exact. Indeed, we have M^~_y - N^~_x = __________ For this exercise we can find an integrating factor which is a function of x alone since M^~_y - N^~_x/N^~ = ________________ can be considered as a function of x alone. Namely we have mu(x) = _____________ Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where M = __________ N = __________ Which is exact since M_y = __________ N_x = ___________ are equal. This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where F(x, y) = ______________.

Explanation / Answer

M=(4ye3x-6),N=(e3x(4x+6y5))

My=(4e3x-0),Nx=(3e3x(4x+6y5)) +(e3x(4+0))

My=4e3x,Nx=12xe3x+18e3xy5+4e3x

My-Nx=4e3x-12xe3x-18e3xy5-4e3x

My-Nx=-3e3x(4x+6y5)

(My-Nx)/N =-3e3x(4x+6y5)/(e3x(4x+6y5))

(My-Nx)/N =-3

integrating factor (x)=e-3x

new equation (e-3x(4ye3x-6))dx+(e-3xe3x(4x+6y5))dy=0

=> (4y-6e-3x)dx+(4x+6y5)dy=0

M=(4y-6e-3x)

N=(4x+6y5)

My=4

Nx=4

Fx=(4y-6e-3x),Fy=(4x+6y5)

F(x,y)=M dx

F(x,y)=(4y-6e-3x) dx

F(x,y)=(4xy+2e-3x)+g(y)

Fy(x,y)=(4x+0)+gy(y)

(4x+0)+gy(y)=(4x+6y5)

=>g(y)=(6y5) dy

=>g(y)=y6

so F(x,y)=4xy +2e-3x +y6

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