Can Separation of Variables Solve My Math Problems?

prace
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\frac{dy}{dx} = e^{3x} \times e^{2y} So \int \frac{dy}{e^{2y}} = \int e^{3x} dx.
 
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but the original problem had e raised to the 3x + 2y. The only way I know to get rid of that is to take the natural log of both sides right? If you do that, you are left with ln (dy/dx) on the left side. How did you get rid of the natural log there?
 
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By the properties of exponents we know that e^{3x+2y} = e^{3x}\times e^{2y}. So we can separate variables without taking the natural log of both sides. In general, a^{n+m} = a^{n}\times a^{m}
 
oOh... :bugeye: I can't believe I didn't see that! sigh... it is those little things that get me all the time.

thank you so much for your help
 
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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