What is the solution for (x+y+4dx)+(2x+2y+1dy)=0 ?

The solution for (x+y+4dx)+(2x+2y+1dy)=0 is y=-x-1+(c_{1})/(e^{3x)}

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(x+y+4dx)+(2x+2y+1dy)=0

(x + y + 4 d x) + (2 x + 2 y + 1 d y) = 0

y = - x - 1 + \frac{c _{1}}{e ^{3 x}}

Solution steps

(x + y + 4 d x) + (2 x + 2 y + 1 d y) = 0

Solve linear ODE:

y = - x - 1 + \frac{c _{1}}{e ^{3 x}}

y = - x - 1 + \frac{c _{1}}{e ^{3 x}}

f (x) = 4 e^{x} + x, (0, 4)

\frac{\partial}{\partial x} (\frac{1}{3} t^{6} - 6 t^{4} + t)

y \frac{d y}{d x} 2 y = 2 + x^{2}

\int arctan (1) d x

x \to 0 lim (x (e^{x} + \frac{1}{x}))

What is the solution for (x+y+4dx)+(2x+2y+1dy)=0 ?
The solution for (x+y+4dx)+(2x+2y+1dy)=0 is y=-x-1+(c_{1})/(e^{3x)}