Find Differential Equation with This Solution

AATroop
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1. Find a differential equation with the solution y(x) = (x + C)3

The answer cannot contain C)



2. There are no relevant equations.


3. I'm not entirely sure how to do this; I understand that a differential equation has multiple solutions, but for some reason I'm lost on how to find the equation from the solution itself.
 
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AATroop said:
1. Find a differential equation with the solution y(x) = (x + C)3

The answer cannot contain C)



2. There are no relevant equations.


3. I'm not entirely sure how to do this; I understand that a differential equation has multiple solutions, but for some reason I'm lost on how to find the equation from the solution itself.

Differentiate the solution. See if you can find a way to express x+C in terms of y(x).
 
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Dick said:
Differentiate the solution. See if you can find a way to express x+C in terms of y(x).

Thanks for helping me. I was working on it and I reached \frac{dy}{dx} = 3\times y^{2/3}. I found the derivative and basically just substituted.
 
AATroop said:
Thanks for helping me. I was working on it and I reached \frac{dy}{dx} = 3\times y^{2/3}. I found the derivative and basically just substituted.

And that is exactly correct. If you solve that by separation of variables, you'll get y=(x+C)^3, yes? It's worth a quick check.
 
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Dick said:
And that is exactly correct. If you solve that by separation of variables, you'll get y=(x+C)^3, yes? It's worth a quick check.

Yes, I do. Awesome. Hey, if you don't mind, could you just briefly look over the next question? It's the same format, but y(x) = Cx^3 and I concluded \frac{dy}{dx} = \frac{3\times y}{x}. I just want to make sure that one is correct because the next few questions rely on it.

Thanks again.
 
AATroop said:
Yes, I do. Awesome. Hey, if you don't mind, could you just briefly look over the next question? It's the same format, but y(x) = Cx^3 and I concluded \frac{dy}{dx} = \frac{3\times y}{x}. I just want to make sure that one is correct because the next few questions rely on it.

Thanks again.

You're welcome. Yes, that's a correct form. Check it the same way as the last one. Separate variables and solve it.
 
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Dick said:
You're welcome. Yes, that's a correct form. Check it the same way as the last one. Separate variables and solve it.

OK, I will. Thank you very much for your help, glad to know I'm finally getting this.
 
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