- #1
BarackObama
- 13
- 0
Homework Statement
dz/dt + e^(t+z) = 0
Homework Equations
The Attempt at a Solution
dz/dt = -e^te^z
integral(dz/e^z) = integral(-e^tdt)
let u = 1/e^z
dv = dz
du = -e^-zdz v= z
integral(udv)
= z/e^z + integral(ze^-zdz)
Another Separable Differential Equation
- Thread starter BarackObama
- Start date
C).In summary, we solved for dz/dt and found that z = -ln(e^t + C) by taking the integral of e^-zdz and using ln to bring the variables down. This allows us to simplify the original equation and find the solution for z.
dz/dt + e^(t+z) = 0
dz/dt = -e^te^z
integral(dz/e^z) = integral(-e^tdt)
let u = 1/e^z
dv = dz
du = -e^-zdz v= z
integral(udv)
= z/e^z + integral(ze^-zdz)
- #2
tiny-tim
Science Advisor
Homework Helper
- 25,839
- 258
Welcome to PF!
Good morning, Mr President! Welcome to PF!
(have an integral: ∫ and try using the X2 icon just above the Reply box)
So far so good …
Now just rewrite ∫ dz/ez as ∫ e-z dz = … ?
Good morning, Mr President! Welcome to PF!
(have an integral: ∫ and try using the X2 icon just above the Reply box
)BarackObama said:
So far so good …
Now just rewrite ∫ dz/ez as ∫ e-z dz = … ?
- #3
BarackObama
- 13
- 0
Good morning!
∫e^-zdz = ∫-e^tdt
-e^-z = -e^t + C
e^-z = e^t + C
Is there any way to bring the variables down?
Thanks!
∫e^-zdz = ∫-e^tdt
-e^-z = -e^t + C
e^-z = e^t + C
Is there any way to bring the variables down?
Thanks!
- #4
Mark44
Mentor
- 37,660
- 9,916
Take ln of both sides, but keep in mind that the log of a sum doesn't simplify.
- #5
BarackObama
- 13
- 0
z = -ln(e^t + C)
... not exactly easy to verify
... not exactly easy to verify
- #6
Mark44
Mentor
- 37,660
- 9,916
What's so hard about it? Note the -ln(e^t + C) = ln(1/(e^t + C))
Related to Another Separable Differential Equation
1. What is a separable differential equation?
A separable differential equation is a type of differential equation where the variables can be separated and solved independently. This means that the equation can be rewritten in the form of dy/dx = f(x)g(y), where f(x) is a function of x and g(y) is a function of y.
2. How do you solve a separable differential equation?
To solve a separable differential equation, you first rearrange the equation so that all terms with y are on one side and all terms with x are on the other side. Then, you integrate both sides with respect to their respective variables. This will result in an implicit equation, which can be solved for y to obtain the solution.
3. What are some real-world applications of separable differential equations?
Separable differential equations are commonly used in physics, engineering, economics, and other fields to model various physical processes. For example, they can be used to model population growth, radioactive decay, and chemical reactions.
4. Can all differential equations be solved using the separable method?
No, not all differential equations can be solved using the separable method. This method only works for equations that can be written in the form of dy/dx = f(x)g(y). Other types of equations, such as exact equations or non-separable first-order equations, require different methods to solve.
5. Are there any limitations to using the separable method for solving differential equations?
One limitation of the separable method is that it only works for first-order differential equations. It is also not always possible to find an explicit solution using this method, and sometimes only implicit solutions can be obtained. Additionally, the separable method may not be applicable to certain special cases of differential equations, such as singular or degenerate equations.
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