Equation of Curve Passing Through (0, 9)

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Homework Statement



A curve passes through the point (0, 9) 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?

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The Attempt at a Solution



im thinking the slope would then be 18...but i haven't gotten much farther than that
 
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Keep in mind that derivative equals slope at a given point.
 
EstimatedEyes said:
Keep in mind that derivative equals slope at a given point.

Right. And slope=derivative. So you want to solve the differential equation y'(x)=2*y(x).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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