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

d er i v a t i v e \frac{3}{x ^{4} + 7 x + 1}

x \to 1 lim ([2 f (x) + g (x)])

t an g e n t y = (\frac{( x + 2 )}{( x - 2 ) ^{2}})

\int \frac{- 1}{( x - 1 ) ^{2}} d x

y^{^{''}} + 6 y^{^{'}} + 9 = e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

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