Solve Initial Value Problem y'=y^2: Solution & Interval Determination

In summary: I'll definitely be looking into this more and trying to get a better understanding.In summary, the problem is that the solution to the differential equation is only valid for a certain interval (- \infty, 1), but when t=1, the y(x) equation becomes undefined, so the other interval (- \infty, 1) should be included as a valid solution.
  • #1
2RIP
62
0

Homework Statement


Solve the initial value problem
[tex]y'=y^2 , y(0)=1[/tex] and determine the interval in which the solution exists.

The Attempt at a Solution


So after solving the differential equation, we get
[tex] y= 1/(1-t) [/tex] However, I don't understand why the interval is only [tex] (- \infty , 1 ) [/tex] and not [tex](- \infty, 1 ) U ( 1 , \infty)[/tex]. From the y(x) equation, we see that it is only not defined when t=1, so why don't we include the other interval as a solution? Could someone please enlighten me please.
 
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  • #2
You think of the differential equation as describing how the system evolves starting from t=0 to another value of t. You can't reach t in (1,inf) without going through the singularity at t=1.
 
  • #3
2RIP said:

Homework Statement


Solve the initial value problem
[tex]y'=y^2 , y(0)=1[/tex] and determine the interval in which the solution exists.

The Attempt at a Solution


So after solving the differential equation, we get
[tex] y= 1/(1-t) [/tex] However, I don't understand why the interval is only [tex] (- \infty , 1 ) [/tex] and not [tex](- \infty, 1 ) U ( 1 , \infty)[/tex]. From the y(x) equation, we see that it is only not defined when t=1, so why don't we include the other interval as a solution? Could someone please enlighten me please.

Well, when you graph the differential equation y'=y^2, the function of course only exists above the x-axis (because y^2 will always make any y value positive). Therefore, when you solve the differential equation and get y=1/(1-t), any t value you plug may or may not make the y= value positive/negative; however, the slope will always be positive, so the graph should only go from [tex] (- \infty , 1 ) [/tex]. Note the asymptotes at x=1 and y=0, anything to the right of x=1 will be negative but the differential (y^2) will only account for positive values.
 
  • #4
Oh, so the value t=0 which was part of the initial condition y(0)=1, must be contained in a valid interval until there is discontinuity on either ends or goes to +/-infinity?

Thanks so much guys for your valuable inputs.
 

Related to Solve Initial Value Problem y'=y^2: Solution & Interval Determination

1. What is an Initial Value Problem?

An Initial Value Problem (IVP) is a type of mathematical problem that involves finding a function that satisfies a given set of conditions, typically at a single point. These conditions include an initial value for the independent variable, as well as one or more initial values for the dependent variable(s). The goal of an IVP is to find a solution that satisfies both the equation and the initial values.

2. What are the key components of an Initial Value Problem?

The key components of an Initial Value Problem are the differential equation, the initial value(s), and the solution. The differential equation represents the relationship between the independent and dependent variables, while the initial value(s) provide a starting point for the solution. The solution is the function that satisfies both the equation and the initial value(s).

3. How is an Initial Value Problem different from a Boundary Value Problem?

An Initial Value Problem is different from a Boundary Value Problem in that an IVP requires initial values for the dependent variable(s), while a BVP requires boundary values for the dependent variable(s). Boundary values are values of the dependent variable(s) at specific points in the domain, while initial values are values at a single point in the domain. Additionally, solutions to IVPs are typically unique, while solutions to BVPs may have multiple solutions.

4. What are some common methods for solving Initial Value Problems?

Some common methods for solving Initial Value Problems include the Euler method, the Runge-Kutta methods, and the Taylor series method. These methods involve using numerical approximations to find an approximate solution to the IVP. Other methods include separation of variables, substitution, and variation of parameters, which involve manipulating the equation in order to find an exact solution.

5. Why are Initial Value Problems important in science?

Initial Value Problems are important in science because they allow us to model and understand real-world phenomena. Many physical, biological, and chemical processes can be described using differential equations, and IVPs allow us to find solutions that match real-world data. These solutions can then be used to make predictions and inform further scientific research and experimentation.

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