# Find a particular solution of the differential equation , given that y = 0 when x = 0.

It is given that

On integrating both sides, we get,

----------------(1)

Let

eydt = -dt

Substituting value in equation (1), we get,

-log|t| = log|C(x+1)|

-log|2 – ey| = log|C(x + 1)|

------------------(2)

Now, at x = 0 and y = 0, equation (2) becomes,

C = 1

Now, substituting the value of C I equation (2), we get,

Therefore, the required particular solution of the given differential equation is

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