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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! $\sqrt{7x^3} = \sqrt{7}x^{3/2}$. –  Arturo Magidin Jul 19 '12 at 3:07
    
I have to use the substitution for the 7x^3 –  user69 Jul 19 '12 at 3:10
    
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. –  Arturo Magidin Jul 19 '12 at 3:12
    
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. –  Arturo Magidin Jul 19 '12 at 3:23

3 Answers 3

up vote 1 down vote accepted

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. –  user69 Jul 19 '12 at 3:25
    
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? –  user69 Jul 19 '12 at 3:30
1  
@Kudla69 the $dx$ is just to show what variable you are integrating with respect to (in this case, $x$). –  Joe Jul 19 '12 at 3:36

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

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