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

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

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(dy)/(dx)-y/(x+1)=0

\frac{d y}{d x} - \frac{y}{x + 1} = 0

y = e^{c_{1}} x + e^{c_{1}}

Solution steps

\frac{d y}{d x} - \frac{y}{x + 1} = 0

Solve separable ODE:

y = e^{c_{1}} x + e^{c_{1}}

y = e^{c_{1}} x + e^{c_{1}}

a y^{^{'}} + y = c

(2 y + 2) d y - (4 x^{3} + 6 x) d x = 0

\frac{d x}{d y} = sin (x) (cos (2 y) - cos^{2} (y))

\frac{d}{d θ} (f (θ)) f (θ) = sin (θ) + csc (θ)

4 ln (x) d x - x sec^{2} (y) d y = 0

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