Undetermined coefficients + initial value problem

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Homework Help Overview

The problem involves solving a second-order linear differential equation with constant coefficients, specifically y'' + y' - 2y = 2x, along with initial conditions y(0) = 0 and y'(0) = 1. The original poster is attempting to find the particular solution (yp) after determining the complementary solution (yc).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the distinction between the complementary solution (yc) and the particular solution (yp). The original poster proposes a form for yp and evaluates its derivatives but encounters difficulties. Others suggest alternative forms for yp and discuss the implications of the initial conditions for determining constants in the general solution.

Discussion Status

There is an ongoing exploration of the appropriate form for the particular solution, with some participants providing suggestions for adjustments. The original poster has made progress in finding the general solution and is considering how to apply the initial conditions to solve for the constants.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement for linear independence of the solutions and the need to apply initial conditions effectively. There is an acknowledgment of the complexity involved in determining the particular solution.

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Homework Statement


Find solution for y'' + y' -2y = 2x with initial values of y(0) = 0, y'(0) = 1

Homework Equations


I have found yc = c1*exp(-2x) + c2*exp(x), but finding yp is what I'm having trouble with... AND THEN I'm not so sure how to go about the initial value.


The Attempt at a Solution


Ok, so my "reasonable assumption" :
yp = Ax
y'p = A
y''p= 0

When I plug into equation, I get:
0 + A -2(Ax) = 2x... and that's where I'm stuck.

Any insights?
 
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Is yc the general solution and yp the particular?

If so, shouldn't you be using the initial conditions to solve for c1 and c2?
 
Last edited:
Edit: general solution y = yc + yp.

Does that help?

My strategy is to find the yc part of the general y solution, pretending like the left side of the original eqn is equal to zero. Then to find yp I'm supposed to make a reasonable assumption to find it. All parts of yc and yp must be linearly independent, so yp must be in a similar form of the right side of the original equation. So, I take yp, take 1st and 2nd derivatives and see if whatever I get agrees with 2x. Then when i find what yp is, I add it to yc.

...and that's all I know about this equation.
 
Your "reasonable assumption" is reasonable, but not general enough. Try yp=Ax+B.
 
In general, for a RHS of xn, you will need yp= Axn+ Bxn-1+ Cxn-2 . . .
 
Dick said:
Your "reasonable assumption" is reasonable, but not general enough. Try yp=Ax+B.
Ahhh... Clever.
 
It worked! Thanks much!

Now with the initial values:
I have my general solution y = c1*exp(-2x) + c2*exp(x) -x -1/2

All I'll do to find c1 and c2 is use y(0)=0 in that equation, then take its derivative and use the y'(0)=1, that's the way to go, correct?
 
Last edited:
HallsofIvy said:
In general, for a RHS of xn, you will need yp= Axn+ Bxn-1+ Cxn-2 . . .

Gotcha.
 

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