# Find the equation of the curve? A curve passes through the point (0,2) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? (Use x as the independent variable.) y(x)=______________?

The slope of the tangent line is given by the first derivative. Since the slope is equal to twice the y value we have `(dy)/(dx)=2y` Then:

`(dy)/y=2dx`  Integrating we get:

`lny=2x+C_1`  exponentiating both sides we get:

`e^(lny)=e^(2x+C_1)=e^(2x)e^(C_1)` Let `C=e^(C_1)`

`y=Ce^(2x)`

Since the curve passes through (0,2) we have:

`2=Ce^(2*0)==>C=2`

So the function is `y=2e^(2x)`

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Checking we see that `(dy)/(dx)=4e^(2x)=2y` as required.

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