Solve the initial value problem?

Click For Summary
SUMMARY

The discussion focuses on solving the initial value problem defined by the differential equation (dy/dx) = (1) / (2√x) + 10 with the initial condition y(1) = -1. The solution involves breaking down the problem into two simpler parts: finding functions f(x) and g(x) such that f'(x) = (1) / (2√x) and g'(x) = 10. The approach emphasizes the concept of "differentiating backwards" and suggests using integration to find the original functions, specifically f(x) = ∫(1 / (2√x))dx and g(x) = ∫10dx.

PREREQUISITES
  • Understanding of basic calculus concepts, particularly differentiation and integration.
  • Familiarity with the rules of differentiation and integration.
  • Knowledge of initial value problems in differential equations.
  • Ability to manipulate algebraic expressions and functions.
NEXT STEPS
  • Learn the techniques for solving initial value problems in differential equations.
  • Study the rules of integration, particularly for functions involving square roots.
  • Practice differentiating and integrating various functions to reinforce understanding.
  • Explore applications of differential equations in real-world scenarios.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and differential equations, as well as educators looking for methods to teach these concepts effectively.

femmed0ll
Messages
6
Reaction score
0
1) (dy/dx) = (1) / ( 2√x ) + 10 , y(1) = -1
2) (d^(2)r / dt^2) = ( 4 / t^3 ) ; (dr/dt) | t=1 = -5, r(1)=5

... i have no idea where to start.
what are the steps needed to figure out the answer...can someone please solve this for me! t
thank you so much.
 
Physics news on Phys.org
(1) So the problem here is that you need to find a function y(x) such that when you differentiate it, you get (1) / ( 2√x ) + 10, right? Recall that the operation of differentiating is linear. Meaning that for any two functions f,g, we have (f + g)'=f ' + g'. This observation may help you to simplify the problem. Indeed, if you can find two functions f(x), g(x) such that f '(x)=(1) / ( 2√x ) and g'(x)=10, then the solution to the problem will be y(x) = f(x) + g(x). Do you agree?

Ok, so we've split the difficult problem of finding a function y(x) such that y'(x) = (1) / ( 2√x ) + 10 into the 2 easier problems of finding functions f(x), g(x) such that f '(x)=(1) / ( 2√x ) and g'(x)=10.

This is just a matter of "differentiating backwards". You are used to finding the derivative of a given function. Here, you are given a function and must find it is the derivative of what function.

If you have practiced differentiating enough, then this should not be too hard. Start with finding g(x) as it is somewhat easier than finding f(x). Help yourself with a table listing the rules of differentiation if necessary.

Equivalently, if you know what an integral is and if you're familiar with the rules for solving them, then

[tex]f(x)=\int \frac{1}{2\sqrt{x}}dx[/tex]

and

[tex]g(x)=\int 10dx[/tex]
 
Last edited:

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K