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First order separable differential equation

  • #1
120
0

Homework Statement



x[itex]\frac{dy}{dx}[/itex] = 4y


Homework Equations



I'm not sure if there is a specific equation for these type of problems. My professor just says to separate the two different variables and then integrate them with respect to x.


The Attempt at a Solution



[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] = [itex]\frac{1}{x}[/itex]

∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx =∫[itex]\frac{1}{x}[/itex] dx

left side:

∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx

u = y
du = [itex]\frac{dy}{dx}[/itex] dx

[itex]\frac{1}{4}[/itex]∫[itex]\frac{1}{u}[/itex]du

ln(4y) + [itex]C_{y}[/itex]


right side:

∫[itex]\frac{1}{x}[/itex] dx

ln(x) + [itex]C_{x}[/itex]

ln(4y) = ln(x) + C

Do i have to continue or could I just stop here?

4y = [itex]e^{x}[/itex] * [itex]e^{c}[/itex]

y = [itex]\frac{e}{4}^{x}[/itex] * [itex]\frac{e}{4}^{c}[/itex]
 
Last edited by a moderator:

Answers and Replies

  • #2
33,506
5,191
Please don't use SIZE tags.

Homework Statement



x[itex]\frac{dy}{dx}[/itex] = 4y


Homework Equations



I'm not sure if there is a specific equation for these type of problems. My professor just says to separate the two different variables and then integrate them with respect to x.


The Attempt at a Solution



[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] = [itex]\frac{1}{x}[/itex]

∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx =∫[itex]\frac{1}{x}[/itex] dx

left side:

∫[itex]\frac{1}{4y}[/itex] [itex]\frac{dy}{dx}[/itex] dx

u = y
This is never a good substitution, because all you're doing is switching to a different letter.

du = [itex]\frac{dy}{dx}[/itex] dx

[itex]\frac{1}{4}[/itex]∫[itex]\frac{1}{u}[/itex]du

ln(4y) + [itex]C_{y}[/itex]
You should have ended up with (1/4)ln|y| on the left side.

You made the problem considerably harder than it is by not immediately moving the constant outside the integral, and by dragging dy/dx along.

∫1/(4y)dy = (1/4)∫dy/y
right side:

∫[itex]\frac{1}{x}[/itex] dx

ln(x) + [itex]C_{x}[/itex]

ln(4y) = ln(x) + C

Do i have to continue or could I just stop here?

4y = [itex]e^{x}[/itex] * [itex]e^{c}[/itex]
You have another mistake above. When you exponentiate the right side, you get eln(x) + C = eln(x) * eC. That first factor can be greatly simplfied.
y = [itex]\frac{e}{4}^{x}[/itex] * [itex]\frac{e}{4}^{c}[/itex]
 
  • #3
120
0
Sorry, I didn't know that wasn't allowed.

I was kind of following the example my teacher did that was similar to this problem.

I remember my calc 2 teacher telling me the integral of 1/Ax is ln(Ax). So that's why I thought it was ln(4y). When I did u substitution I thought it was kind of strange how the 4 ends up back inside ln. I guess I heard him wrong


You have another mistake above. When you exponentiate the right side, you get eln(x) + C = eln(x) * eC. That first factor can be greatly simplfied.
Yeah I know that e^ln(x) is just x i just copied it down incorrectly when i was posting it here.


So would it be,

(1/4)ln(y) = ln(x) +C

ln(y) = 4ln(x) +C

y = 4x +C

I don't think I exponentiated this correctly. If it were e^ln(4x) then it would be 4x.

Can e^4(lnx) be written as e^4 * e^ln(x)?

I don't think that's right either because e^(ln(x) +c) is e^ln(x) * e^c. I'm confused.
 
Last edited:
  • #4
33,506
5,191
Sorry, I didn't know that wasn't allowed.
It's not that it isn't allowed - it's just not needed most of the time.
I was kind of following the example my teacher did that was similar to this problem.

I remember my calc 2 teacher telling me the integral of 1/Ax is ln(Ax).
Either your teacher told you something that's wrong or you misheard. You can check this formula by differentiating. d/dx(ln(Ax)) = 1/(Ax) * A = 1/x ≠ 1/(Ax)
So that's why I thought it was ln(4y). When I did u substitution I thought it was kind of strange how the 4 ends up back inside ln. I guess I heard him wrong




Yeah I know that e^ln(x) is just x i just copied it down incorrectly when i was posting it here.


So would it be,

(1/4)ln(y) = ln(x) +C

ln(y) = 4ln(x) +C
Above, it would be 4C, but that's just a different constant. Some people just write C' to indicate that it's different from C. For this problem C' = 4C.
y = 4x +C
No, that's incorrect.

Starting from the equation before this one...

ln(y) = 4ln(x) + C'

eln(y) = e4ln(x) + C' = e4ln(x) * eC'
=> y = eC' * e4ln(x) = Aeln(x4) = Ax4

Note that eC' is just another constant, so call it A.

If you really want to keep track, A = eC' = e4C

Strictly speaking, there should be absolute values involved on both sides. The real formulas are
$$ \int \frac{dy}{y} = ln|y| + C$$
and
$$ \int \frac{dx}{x} = ln|x| + C$$
I don't think I exponentiated this correctly. If it were e^ln(4x) then it would be 4x.

Can e^4(lnx) be written as e^4 * e^ln(x)?
I'm not sure what you mean by e^4(lnx). Is this e^4 * ln(x) or e^(4ln(x))?
I don't think that's right either because e^(ln(x) +c) is e^ln(x) * e^c. I'm confused.
 

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