Question

The equation

$$ \left(3 y e^{x}-2\right) d x+\left(e^{x}\left(3 x+4 y^{3}\right)\right) d y=0 $$

in differential form \(\widetilde{M} d x+\widetilde{N} d y=0\) is not exact. Indeed, we have

$$ \bar{M}_{y}-\widetilde{N}_{x}= $$

For this exercise we can find an integrating factor which is a function of \(x\) alone since

$$ \frac{\bar{M}_{y}-\bar{N}_{x}}{\bar{N}}= $$

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 \(M d x+N d y=0\) where

$$ M= $$

$$ N= $$

Which is exact since

$$ \begin{aligned} &M_{y}= \\ &N_{x}= \end{aligned} $$

are equal.

This problem is exact. Therefore an implicit general solution can be written in the form \(F(x, y)=C\) where

$$ F(x, y)= $$

Answer 1

(3ye^x -2)dx + e^x (3x + 4y63)dy = 0 equation 1 Mdx + Ndy = 0 M = 3ye^x -2 N = e^x(3x + 4y^3) partial differential M/ partial differential y = 3e^x partial differential N/ partial differential x = 3e^x + e^x (3x + 4y^3) it is not exact M_y - N_x/N = f(x) I.F implies mu(n) = e^integral f(x) dx M_y - N_x = 3e^x -3e^x - e^x (3x + 4y^3) M_y - N_x = -e^x (3x + 4y^3) f(x) = M_y - N_x/N = -e^x(3x + 4y^3)/e^x(3x + 4y63) = -1 mu(n) = e^integral -1 dx = e^-x e^-x (3y -2e^-x)dx + (3x + 4y^3)dy = 0 M = 3y -2e^-x N = 3x + 4y^3 M_y = 3 N_x = 3 now it is exact general solution integral_y = c M dx + integral_not x N dy = c integral (3y -2e^-x)dx + integral 4y63 dy = c 3xy + 2e^-x + y^4 = c

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