# Homework Help: Easy Differential Equation Calc 1

1. Oct 1, 2008

### Sheneron

1. The problem statement, all variables and given/known data
Is y = x^2 a solution to the DE $$y'' - 4xy' + 4y = 0$$

3. The attempt at a solution
So:
$$y = x^2$$
$$y' = 2x$$
$$y'' = 2$$

$$2 - 4x(2x) + 4x^2 = 0$$
$$2 - 8x^2 + 4x^2 = 0$$
$$2 - 4x^2 = 0$$

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.

Last edited: Oct 1, 2008
2. Oct 1, 2008

### HallsofIvy

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.

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!

3. Oct 1, 2008

### 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.

4. Oct 1, 2008

### gabbagabbahey

Is $2-4x^2$ zero for all x?!

5. Oct 1, 2008

### 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.

6. Oct 1, 2008

### Sheneron

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

7. Oct 1, 2008

### gabbagabbahey

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

$$y'' - 4xy' + 4y=2-4x^2$$

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

8. Oct 1, 2008

### Sheneron

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

9. Oct 1, 2008

### gabbagabbahey

Not really, for a solution to satisfy the DE it must satisfy it for all values of $x$.

10. Oct 1, 2008

### Sheneron

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

11. Oct 1, 2008

### gabbagabbahey

It's true that $2-4x^2=0$ for $x=1/\sqrt{2}$, but $y(1/\sqrt{2})=1/2$ 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.

12. Oct 1, 2008

### Sheneron

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

13. Oct 1, 2008

### gabbagabbahey

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

14. Oct 1, 2008

### Sheneron

Alright thanks