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