First order separable differential equation

In summary: No, e^(ln(x) + c) = e^ln(x) * e^c = x * e^c. But you should have e^4ln(x) = e^(ln(x4)) = x4.In summary, when solving the differential equation x(dy/dx) = 4y, you must first separate the variables and integrate both sides with respect to their respective variables. This results in ln(4y) = ln(x) + C, where C is a constant. By exponentiating both sides, we get 4y = e^(ln(x)) * e^C, which simplifies to y = Ax^4, where A = e^C. Therefore, the solution to the
  • #1
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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]
 
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  • #2
Please don't use SIZE tags.
november1992 said:

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.

november1992 said:
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
november1992 said:
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.
november1992 said:
y = [itex]\frac{e}{4}^{x}[/itex] * [itex]\frac{e}{4}^{c}[/itex]
 
  • #3
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


Mark44 said:
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.
 
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  • #4
november1992 said:
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.
november1992 said:
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)
november1992 said:
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.
november1992 said:
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
$$ \int \frac{dx}{x} = ln|x| + C$$
november1992 said:
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))?
november1992 said:
I don't think that's right either because e^(ln(x) +c) is e^ln(x) * e^c. I'm confused.
 

1. What is a first order separable differential equation?

A first order separable differential equation is a type of differential equation that can be written in the form of dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y, respectively. This means that the variables x and y can be separated on opposite sides of the equation, making it easier to solve.

2. How do you solve a first order separable differential equation?

To solve a first order separable differential equation, you can follow these steps:

  1. Separate the variables x and y on opposite sides of the equation.
  2. Integrate both sides of the equation with respect to x and y, respectively.
  3. Add a constant of integration to the side of the equation that was integrated with respect to y.
  4. Solve for y in terms of x to get the general solution.
  5. If initial conditions are given, use them to find the specific solution.

3. What are the applications of first order separable differential equations?

First order separable differential equations are used to model a wide range of phenomena in physics, engineering, and other sciences. Some examples of applications include population growth, radioactive decay, chemical reactions, and electrical circuits.

4. Are there any special cases of first order separable differential equations?

Yes, there are two special cases of first order separable differential equations:

  • Homogeneous: In a homogeneous first order separable differential equation, f(x) and g(y) are both functions of the same variable, such as x. This means that the equation can be written in the form dy/dx = F(y/x), where F is a function of y/x.
  • Exact: An exact first order separable differential equation is one that can be written in the form M(x,y)dx + N(x,y)dy = 0, where M and N are functions of x and y that satisfy the condition My = Nx. These equations can be solved using a different method than the one used for general first order separable differential equations.

5. Can all first order differential equations be solved using separation of variables?

No, not all first order differential equations can be solved using separation of variables. Some equations may require other methods, such as substitution or using an integrating factor. Additionally, certain types of first order differential equations, such as nonlinear or partial differential equations, cannot be solved using separation of variables.

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