Easy Differential Equation Calc 1

[Sheneron]

Homework Statement

Is y = x^2 a solution to the DE [tex]y'' - 4xy' + 4y = 0[/tex]

The Attempt at a Solution

So:
[tex]y = x^2[/tex]
[tex]y' = 2x[/tex]
[tex]y'' = 2[/tex]

[tex] 2 - 4x(2x) + 4x^2 = 0[/tex]
[tex] 2 - 8x^2 + 4x^2 = 0[/tex]
[tex] 2 - 4x^2 = 0[/tex]

But doing this I won't get an answer of whether it is a solution or not. I get a variable equal to a number. So what can I do differently? How do I solve this? Thanks in advance.

 

[HallsofIvy]

Homework Statement

Is y = x^2 a solution to the DE [tex]y" - 4xy' + 4y = 0[/tex]

Which means that this must be true for all x.

Suppose you were asked whether x= 2 is a solution to x³- 3x2[/sup+ 10x+ 4= 0. Putting 2 in place of x gives you 2³- 3(2²)+ 10(2)+ 4= 8- 12+ 20+ 4= 20, not 0. Since the equation is not true, 2 is NOT a solution to the problem.

The Attempt at a Solution

So:
[tex]y = x^2[/tex]
[tex]y' = 2x[/tex]
[tex]y" = 2[/tex]

[tex] 2 - 4x(2x) + 4x^2 = 0[/tex]
[tex] 2 - 8x^2 + 4x^2 = 0[/tex]
[tex] 2 - 4x^2 = 0[/tex]But doing this I won't get an answer of whether it is a solution or not. I get a variable equal to a number. So what can I do differently? How do I solve this? Thanks in advance.

Is that true for all x? If yes, then y= x² is a solution to the differential equation. If not, then it is not a solution!

Homework Statement

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Homework Equations

The Attempt at a Solution

 

[Sheneron]

So for any x, y = x^2 is a solution to that DE? And that is because I got a variable equal to a number? And just for a reference the answer in the back of the book said that it wasn't a solution to that DE.

 

[gabbagabbahey]

Is [itex]2-4x^2[/itex] zero for all x?!

 

[Sheneron]

Sorry the post I responded to before didn't load fully and I only saw a small part. Which I was equally confused about.

 

[Sheneron]

So let me get this straight. Since [tex]2 - 4x^2 = 0[/tex] is not true for all x then, [tex] y= x^2[/tex] is not a solution to that DE. If however, that equation had been true for all x it would be a solution?

 

[gabbagabbahey]

Yes, by plugging in [itex] y=x^2[/itex] to the left side of the DE, you showed that:

[tex]y'' - 4xy' + 4y=2-4x^2 [/tex]

which does not equal the right side of the DE (zero) for all [itex]x[/itex], so [itex] y=x^2[/itex] does not satisfy the DE.

 

[Sheneron]

Furthermore, if they had said [tex]x = \sqrt{1/2}[/tex] a solution to the DE the answer would be yes, since that yields a result of 0 = 0?

 

[gabbagabbahey]

Not really, for a solution to satisfy the DE it must satisfy it for all values of [itex]x[/itex].

 

[Sheneron]

with x at that value, and y the same as above, then y would be a solution to that DE

 

[gabbagabbahey]

It's true that [itex]2-4x^2=0[/itex] for [itex]x=1/\sqrt{2}[/itex], but [itex]y(1/\sqrt{2})=1/2[/itex] which is a constant, so dy/dx and d^2y/dx^2 are both zero and your left with 4*(1/2)=0 which is false. It is meaningless to say that y(a certain value) satisfies the DE, since y(a certain value) is a value, not a function. Only functions satisfy DE's.

 

[Sheneron]

Forget that y part, I didn't mean to put that. If I said is [tex]y = x^2 [/tex] at [tex]x = 1/\sqrt{2}[/tex] a solution to that DE what would the answer be?

 

[gabbagabbahey]

Still no; [itex]y=x^2[/itex] at [itex]x=1/\sqrt{2}[/itex] is just a value not a function. Functions satisfy DE's, values don't.

 

[Sheneron]

Alright thanks