Differential Equation

reallreality said:
View attachment 38138
Determine if y_1 and y_2 are solutions to the differential equation.
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A function is a solution to a differential equation if it makes the equation true (for all x).

So, for each of these two functions ([imath]y_1[/imath] and [imath]y_2[/imath]), find the second derivative, and see if it is equal to 9 times the function itself.

What do you find? Please show your work, so we can guide you if you need more help.
 

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reallreality said:
View attachment 38138
Determine if y_1 and y_2 are solutions to the differential equation.
Click to expand...
You can also rearrange the differential equation like this:

[imath]y'' - 9y = 0[/imath]

And combine the two solutions like this:

[imath]y = y_1 + y_2 = e^{3x} + e^{-3x}[/imath]

Now you need to find [imath]y''[/imath], then plug [imath]y''[/imath] and [imath]y[/imath] in the rearranged differential equation. If you get zero, they are solutions to the differential equation.
 
mario99 said:
And combine the two solutions like this:

[imath]y = y_1 + y_2 = e^{3x} + e^{-3x}[/imath]

Now you need to find [imath]y''[/imath], then plug [imath]y''[/imath] and [imath]y[/imath] in the rearranged differential equation. If you get zero, they are solutions to the differential equation.
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Technically, of course, this would only show that the sum [imath]y_1+y_2[/imath] is a solution, not that each function is individually, which is required. The student will eventually learn that the latter implies the former, though not vice versa.
 
Dr.Peterson said:
Technically, of course, this would only show that the sum [imath]y_1+y_2[/imath] is a solution, not that each function is individually, which is required. The student will eventually learn that the latter implies the former, though not vice versa.
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If the sum is a solution, it is guaranteed each term in the sum is also a solution. If you don't agree with this, give me an example that destroys this fact!
 
mario99 said:
If the sum is a solution, it is guaranteed each term in the sum is also a solution. If you don't agree with this, give me an example that destroys this fact!
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Have you ever heard of non-linear differential equations?

-Dan
 
mario99 said:
If the sum is a solution, it is guaranteed each term in the sum is also a solution. If you don't agree with this, give me an example that destroys this fact!
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I'm not an expert in differential equations, so I'm not arguing with you; but this surprises me. If [imath]e^{3x}=(e^{3x}-x)+(x)[/imath] is a solution, then must [imath](e^{3x}-x)[/imath] and [imath](x)[/imath] both be solutions? I'm guessing there is some condition that you haven't mentioned.

Can you refer me to the theorem you are using, so I can be aware of it?

In any case, it seems very likely that this is a beginning problem (about the very meaning of solutions), not one for which theorems about linear differential equations are intended to be used, or likely to be known. Quite possibly, the fact that the sum of known solutions is a solution is going to be taught soon. But I could be wrong.
 
Dr.Peterson said:
I'm not an expert in differential equations, so I'm not arguing with you; but this surprises me. If [imath]e^{3x}=(e^{3x}-x)+(x)[/imath] is a solution, then must [imath](e^{3x}-x)[/imath] and [imath](x)[/imath] both be solutions? I'm guessing there is some condition that you haven't mentioned.

Can you refer me to the theorem you are using, so I can be aware of it?

In any case, it seems very likely that this is a beginning problem (about the very meaning of solutions), not one for which theorems about linear differential equations are intended to be used, or likely to be known. Quite possibly, the fact that the sum of known solutions is a solution is going to be taught soon. But I could be wrong.
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I will not pretend to be better than you Dr. Dave, but your example is invalid (because it is still [imath]e^{3x}[/imath]). Since we're talking about linear differential equations, both ways implies to the other it's a solution. I'm not an expert in theorems. And I think that Dan is an expert in this field and his reply implies what you are talking about here

Dr.Peterson said:
Technically, of course, this would only show that the sum [imath]y_1+y_2[/imath] is a solution, not that each function is individually, which is required. The student will eventually learn that the latter implies the former, though not vice versa.
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applies for non-linear differential equations or something like that.

I apologize to you Dr.Peterson if this sounds like a challenge. It's not and I'm sure that you are one of the best mathematicians that ever exists in this universe, and an amateur like Mario doesn't stand a chance against you.
 
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