What went wrong with my calculations for a 2nd order D.E. solution?

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In summary, the conversation discusses a problem with finding the correct solution for a differential equation, specifically y``+y`+9.25y=2+2x+x^2. The individual has provided their calculations and solution, but it is noted that the homogeneous solution is not actually a homogeneous solution. They will recheck their work.
  • #1
asdf1
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for the following question:
y``+y`+9.25y=2+2x+x^2

my problem:
yh=c1+c2e^(-x)
suppose yp=c3x^2 + c4x+c5
then yp`=2c3x+c4
so yp``=2c3

then 2c3+2c3x+c4=2+2X+X^2
so c3=1, c4=1
so yp=x^2+x
then y=c1+c2e^(-x)+x^2+x
which implies c1=8
=> y=8+c2e^(-x)+x^2+x
so y`=-c2e^(-x)+2x+1
so -1=-c2+1 =>c2=2
then y=8+2e^(-x)+x^2 +x

but the correct answer should be 3e^(-x)+5+2x+(1/3)x^3
does anybody know what's wrong with the calculations?
 
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  • #2
asdf1 said:
for the following question:
y``+y`+9.25y=2+2x+x^2
my problem:
yh=c1+c2e^(-x)

According to what you have written your homogeneous solution isn't a homogeneous solution:

(c_2e^(-x))+(-c_2e^(-x))+9.25(c_1+c_2e^(-x))=9.25(c_1+c_2e^(-x)) which isn't zero (unless both constants are zero)
 
  • #3
jeepers! i'll recheck that again! thanks!
 

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function and its first and second derivatives. It models the relationship between a quantity and its rate of change.

2. How do you solve a second order differential equation?

There are several methods for solving a second order differential equation, including separation of variables, substitution, and using an integrating factor. The method used depends on the specific form of the equation.

3. What are the initial conditions for a second order differential equation?

The initial conditions for a second order differential equation are the values of the function and its first derivative at a given point. These conditions are necessary to find a unique solution to the equation.

4. Can a second order differential equation have more than one solution?

Yes, a second order differential equation can have multiple solutions. This is because the equation may have different initial conditions or the equation itself may have multiple solutions. However, a unique solution can be determined if the necessary initial conditions are provided.

5. How are second order differential equations used in science?

Second order differential equations are used in various fields of science to model real-world phenomena, such as motion, growth, and decay. They are also used in engineering to design and analyze systems and processes. Additionally, they are used in physics and chemistry to describe the behavior of physical systems.

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