Help solving the Initial Value Problem

In summary, the conversation discusses an attempt to solve a second-order differential equation with initial conditions. Two methods are used, one factoring the auxiliary equation and the other using a repeated root. The first method is verified to be correct through WolframAlpha, but the second method results in an extra "t" term in the second solution. This is explained as a result of the repeated root, and the general solution is given as a combination of exponential and power functions.
  • #1
middieman147
8
0

Homework Statement


y''-4y'+4y=0 , y(1)=1 and y'(1)=1

The Attempt at a Solution



Auxiliary equation: r2-4r+4=0

I tried factoring 2 different ways:
(r-2)2=0

r=2,r=2

y1=e2t
y2=y1

y(t)=c1e2t+c2e2t

y(1)=c1e2+c2e2=1 ---eq(1)y'(t)=2c1e2t+2c2e2t

...c2=1/(2e2)-c1 ---eq(2)

sub eq(2) into eq(1) --> 1/2=1method 2:

r(r-4)=-4

r=-4, r=0

y(1)=c1e-4+c2=1 ---eq(1)

y'(1)=c1=-1/(4e-4)

sub eq(2) into eq(1) --> c2=5/4

solution equation:

y(t)=1/4(-1+5)
=1... which doesn't seem correct

i checked the solutions on wolframalpha and it solves it using the first method, but ends up with:

y1=c1e2t
y2=c2e2t *t

can someone explain to me why the extra t shows up in y2? Thanks.

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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  • #2
hi middieman147 :smile:

(i don't understand your second method at all :confused:)
middieman147 said:
i checked the solutions on wolframalpha and it solves it using the first method, but ends up with:

y1=c1e2t
y2=c2e2t *x

can someone explain to me why the extra x shows up in y2?

it isn't an x, it's a t

when a root b is repeated (in the auxiliary equation), the two independent solutions are ebt and tebt

if it's repeated n times, the n independent solutions are ebt tebt … tn-1ebt
 

1. What is an Initial Value Problem (IVP)?

An Initial Value Problem is a mathematical problem that involves finding a function or a set of functions that satisfy a given differential equation and its initial conditions. The initial conditions refer to the values of the function(s) at a specific point in the domain.

2. How do you solve an Initial Value Problem?

There are several methods for solving an Initial Value Problem, such as the Euler method, Runge-Kutta method, and the Picard's method. These methods involve finding a numerical approximation of the solution to the differential equation by evaluating the function(s) at specific points in the domain.

3. What is the importance of solving an Initial Value Problem?

Solving an Initial Value Problem is crucial in various fields of science and engineering, such as physics, chemistry, and economics. It allows us to model real-world phenomena and predict their behavior over time. It also helps in understanding systems and making informed decisions.

4. Can an Initial Value Problem have multiple solutions?

No, an Initial Value Problem can only have a unique solution. This is because the initial conditions uniquely determine the solution to the differential equation. However, there can be different methods of solving the problem, which may result in slightly different solutions.

5. How do you check if a solution to an Initial Value Problem is correct?

To check if a solution to an Initial Value Problem is correct, you can substitute the solution into the original differential equation and initial conditions. If the solution satisfies the equation and initial conditions, then it is a valid solution. You can also use graphical methods, such as slope fields, to visualize the solution and see if it matches the expected behavior.

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