First order separable differential equation

[november1992]

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]

 

[Mark44]

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 e^{ln(x) + C} = e^ln(x) * e^C. That first factor can be greatly simplfied.

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

 

[november1992]

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 e^{ln(x) + C} = e^ln(x) * e^C. 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.

 

[Mark44]

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'

e^ln(y) = e^{4ln(x) + C'} = e^4ln(x) * e^C'
=> y = e^C' * e^4ln(x) = Ae^ln(x4) = Ax⁴

Note that e^C' is just another constant, so call it A.

If you really want to keep track, A = e^C' = e^4C

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.