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

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

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

e^{x} (y - 1) d x + 2 (e^{x} + 4) d y = 0

y = \frac{c _{1}}{e ^{x} + 4} + 1

Solution steps

e^{x} (y - 1) d x + 2 (e^{x} + 4) d y = 0

Solve separable ODE:

y = \frac{c _{1}}{e ^{x} + 4} + 1

y = \frac{c _{1}}{e ^{x} + 4} + 1

x, 4

\frac{3}{x ^{4} + 7 x + 1}

y^{^{'}} = 3 x y

\frac{7 + 2 x}{1 - 6 x}

\frac{x ^{2} - 6 x - 1}{( x - 3 ) ^{2}}

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