Find the particular solution of the equation: e^xdy/dx=4 given that y=3 when x=0?

Let's start by separating the variables: dy = (4 / e^x) dx

Now, we can integrate both sides of the equation: ∫ dy = ∫ (4 / e^x) dx

Integrating, we get: y = ∫ (4 / e^x) dx

To integrate 4 / e^x, we can use the fact that the integral of e^x is e^x:
y = -4e^(-x) + C

Now, we can apply the initial condition y = 3 when x = 0 to determine the value of the constant C:
3 = -4e^(-0) + C
3 = -4(1) + C
3 = -4 + C
C = 7

Therefore, the particular solution of the given equation, satisfying the initial condition, is:
y = -4e^(-x) + 7.
 
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