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$$\frac{dy}{dx} =\sqrt{7x^3}$$

I need to use substitution on the $7x^3$ but I'm a little stuck. My car broke down on the way to class and I missed the lesson!

I get this far..

$$u =7x^3$$ $$du = 21x^2 dx$$ $$dx = \frac{du}{21x^2}$$ $$dy = \sqrt{u} \frac{du}{21x^2}$$

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    The equation is already separated: just integrate! $\sqrt{7x^3} = \sqrt{7}x^{3/2}$.2012-07-19
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    I _have_ to use the substitution for the 7x^32012-07-19
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    *What* substitution? You say "the" substitution, as if there were only one. Do you have a worked out example to show what you mean? What you say makes no sense to me at all.2012-07-19
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    When you do a substitution, you can't have **both** the old and the new variable left over. Your third and fourth line are nonsense. And the "answer in the back of the book" cannot refer to the problem of solving the differential equation $\frac{dy}{dx}=\sqrt{7x}$, because that is *not* a solution to the differential equation. Please copy the problem *verbatim*, including instructions.2012-07-19

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We can use the technique of separation of variables. The method is as follows:

$$dy=\sqrt{7x^3}dx$$ $$\int dy = \int\sqrt{7x^3}dx$$

The justification of this method is a little more nuanced than simply "multiplying" $dx$ and then integrating--we are taking the antiderivative of both sides: $\int\frac{dy}{dx} dx = \int f'(x) dx = f(x)+C$

So $$y = \sqrt{7}\int x^{3/2} dx$$ Thus, we have $$y = \frac{2\sqrt{7}}{5}x^2\sqrt{x}+C$$ for some constant $C$. This is the general solution to this differential equation.

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    thanks! That's the answer that it shows in my book.2012-07-19
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    what happens to the dx at the end? is the antiderivative of dx just C? or is the dx just there to show that its already the differential?2012-07-19
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    @Kudla69 the $dx$ is just to show what variable you are integrating with respect to (in this case, $x$).2012-07-19
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$$dy = \sqrt{7}x^{3/2} dx$$ so $$\int dy = \sqrt{7}\int x^{3/2} dx + K$$ therefore $$y = \frac{2\sqrt{7}}{5}x^{5/2} + K$$ $K$ is the constant of integration. Notice that $\int x^{\alpha}dx = \frac{x^{\alpha+1}}{\alpha+1} + K, \alpha \in \mathbb{R-\{-1\}}$

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$$\frac{dy}{dx}=\sqrt 7\,x^{3/2}\Longrightarrow \int dy=\sqrt 7\,\int x^{3/2}\Longrightarrow y=\frac{2\,\sqrt 7}{5}x^{5/2}+C$$