Solve Differential eq with initial conditions

In summary, the conversation discusses solving the equation for H(t) with a given initial condition, using a Taylor series approach and separating the variables. The initial attempt using Taylor series expansion is shown to be incorrect, but the correct solution is found using separation of variables.
  • #1
billiards
767
16

Homework Statement



Solve: [tex]\frac{\partial H}{\partial t} = -4\kappa H^{2}[/tex]

With initial condition: [tex]H(0) = 1/L^{2}[/tex]

To find: [tex]H(t) = \frac{1}{4\kappa t + L^{2}}[/tex]

2. The attempt at a solution

I tried using Taylor series expansion such that:

[tex]H(t)\approx H(0)+t\frac{\partial H}{\partial t}(0)+...[/tex]

To first order this yielded: [tex]H(t)=\frac{L^{2}-4\kappa t}{L^{4}}[/tex]

This is wrong unless t and/or k equals zero. Therefore this is wrong, it is not the general solution. Please help if you can. Thanks.
 
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  • #2
hi billiards! :smile:
billiards said:
I tried using Taylor series expansion …

eugh!

just separate the variables: dH/H2 = -4k dt :wink:
 
  • #3
tiny-tim said:
separate the variables

Thanks tiny-tim, with that hint I solved it straight away. You have no idea how long I was stuck.

Out of interest, what was wrong with my Taylor series approach?
 
  • #4
billiards said:
Thanks tiny-tim, with that hint I solved it straight away. You have no idea how long I was stuck.

he he :biggrin:
Out of interest, what was wrong with my Taylor series approach?

nothing … they are the same to first order …

what is the inverse of 1 + (4k/L2)t, to first order ? :wink:
 

FAQ: Solve Differential eq with initial conditions

What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is commonly used to model dynamic systems in various fields, including physics, engineering, and economics.

How do you solve a differential equation?

There are various methods for solving differential equations, depending on the type and complexity of the equation. Some common techniques include separation of variables, substitution, and using integrating factors. In some cases, numerical methods or computer software may also be used.

What are initial conditions in a differential equation?

Initial conditions refer to the values of the dependent variable and its derivatives at a specific starting point in the domain of the equation. These values are necessary for solving a differential equation, as they act as constraints on the possible solution.

Can any differential equation be solved with initial conditions?

No, not all differential equations can be solved with initial conditions. Some equations may have no analytical solution, while others may require additional information or assumptions to be solved.

Why are initial conditions important in solving differential equations?

Initial conditions are important because they help to determine the unique solution to a differential equation. Without these conditions, the equation may have multiple solutions or no solutions at all.

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