Consider the following differential equation

$(1+y)\frac{dy}{dx}=y$

The solution of the equation that satisfies the condition y(1) = 1 is

The correct option is ye^y = e^x.

To find the solution to the given differential equation, we start with the equation:
$$(1+y)\frac{dy}{dx}=y$$

We can solve this differential equation by using the method of separation of variables. Let's rearrange the terms to group all y-terms on one side and x-terms on the other side:
$$\frac{1+y}{y}dy=dx$$

Simplifying the fraction on the left-hand side gives:
$$(\frac{1}{y}+1)dy=dx$$

Now, we integrate both sides of the equation:
$$\int (\frac{1}{y}+1)dy=\int dx$$

Integrating term by term, we get:
$$\ln |y|+y=x+C$$
where C is the constant of integration.

We are given the initial condition y(1) = 1, which means y = 1 when x = 1. Let's substitute these values into our integrated equation to solve for C:
$$\ln (1)+1=1+C$$

Since ln(1) = 0, the equation simplifies to:
$$0+1=1+C\Rightarrow C=0$$

Substituting C = 0 back into the general solution, we obtain:
$$\ln (y)+y=x$$

To convert this equation into the form of the options, we exponentiate both sides (raise e to the power of both sides):
$$e^{\ln (y)+y}=e^x$$

Using the laws of exponents, we can rewrite the left-hand side:
$$e^{\ln (y)}\cdot e^y=e^x$$

Since e^ln(y) = y, we get:
$$y{e}^y=e^x$$

This matches the correct option.