What is the solution to solving a linear differential equation?

In summary: Now, what is u(x)? [Hint: u'(x)= u(x) so (u(x)/ u(0))'= u'(x)/ u(0)- u(x)u'(0)/ u(0)^2= u(x)/ u(0)- 0 so that u(x)/ u(0)= e^x]In summary, the conversation discusses using the Euler midpoint method and integrating factor method to approximate and find the exact value of y(0.1) with a step size of 0.1. It also mentions determining the global error with a minimum of 5 decimal places. The equation being used is y'= x+y, y(0) = 1.35. The
  • #1
ariffinaldo
2
0

Homework Statement



1. To use Euler midpoint method to find an approx value to y(0.1) using a step size 0.1
2. To use integrating factor method to find exact value of y(0.1)
3. Determine the global error ( min 5 decimal places)

Homework Equations



f(x,y) = x+y, y(0) = 1.35

The Attempt at a Solution



I have done euler midpoint part which is
y1 = y0 + hf(x0, y0)= 1.35 +0.1(0+0.1) = 1.45
Is my working and answer correct. Pls any help..
 
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  • #2
ariffinaldo said:

Homework Statement



1. To use Euler midpoint method to find an approx value to y(0.1) using a step size 0.1
2. To use integrating factor method to find exact value of y(0.1)
3. Determine the global error ( min 5 decimal places)

Homework Equations



f(x,y) = x+y, y(0) = 1.35

The Attempt at a Solution



I have done euler midpoint part which is
y1 = y0 + hf(x0, y0)= 1.35 +0.1(0+0.1)
This is incorrect. It should be y(0)+ h(x+ y)= 1.35+ .1(0+ 1.35), not "(0+ .1)"

= 1.45
Is my working and answer correct. Pls any help..
 
  • #3
HallsofIvy said:
This is incorrect. It should be y(0)+ h(x+ y)= 1.35+ .1(0+ 1.35), not "(0+ .1)"

Ohh yes.. how could i miss that? How abt the intergrating factor part.. I am a loss for that.
 
  • #4
The differential equation y'= x+y or y'+ y= x is linear. You should have learned a formula for the integrating factor for any linear d.e.

If not, you can work it out as follow: an integrating factor is a function u(x) such that multiplying the equation by it makes the left side a single derivative:
u(x)(y'+ y)= u(x)y'+ u(x)y= (uy)'. By the product rule, (uy)'= uy'+ u'y so that will be equal to uy'+ uy if and only if u'= u. What function satifies that?

Once you have found u, and multiplied by it, your equation will be (uy)'= ux. Integrating that gives
[tex]u(x)y(x)= \int u(x)x dx+ C[/tex]
(The "u(x)" on the right is inside the integration. You cannot just divide both sides by u(x) and eliminate that.)
 

1. What is an Initial Value Problem?

An Initial Value Problem (IVP) is a type of differential equation that involves finding the unknown function given an initial value, or starting point. The initial value is typically a point at which the function is known, and the differential equation describes how the function changes over time.

2. How is an Initial Value Problem solved?

An IVP can be solved using various techniques, such as separation of variables, substitution, or using an integrating factor. The solution will depend on the type of differential equation and the initial conditions given.

3. What is the importance of Initial Value Problems in science?

IVPs are important in science because they allow us to model and understand how systems change over time. They are commonly used in physics, engineering, and other fields to describe the behavior of physical systems.

4. Can an Initial Value Problem have multiple solutions?

Yes, an IVP can have multiple solutions. This is known as an "ill-posed" problem, where the initial conditions are not enough to uniquely determine the solution. In these cases, additional constraints or information may be needed to find a unique solution.

5. What are some real-world applications of Initial Value Problems?

IVPs have many real-world applications, such as modeling population growth, predicting the spread of diseases, and understanding the behavior of electrical circuits. They are also used in fields like economics and finance to model and predict market trends and behavior.

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