# 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...

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)=______________?

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Expert Answers

embizze | Certified Educator

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.