# Undetermined coefficients + initial value problem

• becca4
In summary, the conversation discusses finding the solution for y'' + y' - 2y = 2x with initial values of y(0) = 0 and y'(0) = 1. The general solution yc = c1*exp(-2x) + c2*exp(x) is provided, but the particular solution yp is still being determined. The strategy is to find the yc part of the general solution and then make a reasonable assumption to find yp. The "reasonable assumption" initially made is not general enough, so it is adjusted to yp = Ax + B. With the new assumption, the particular solution is found and added to yc to get the general solution. Finally, the initial values are used to solve
becca4

## 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?

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.

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.

## 1. What is the "undetermined coefficients" method?

The undetermined coefficients method is a technique used in solving differential equations, specifically for solving non-homogeneous equations with constant coefficients. It involves finding a particular solution by assuming a solution in the form of the non-homogeneous term and solving for the undetermined coefficients.

## 2. How is the undetermined coefficients method used in initial value problems?

In initial value problems, the undetermined coefficients method is used to find a particular solution that satisfies both the differential equation and the given initial conditions. The method involves solving for the coefficients using the initial conditions and then combining them with the complementary solution.

## 3. What are the limitations of the undetermined coefficients method?

One limitation of the undetermined coefficients method is that it can only be used for equations with constant coefficients. It is also not applicable for all types of non-homogeneous terms, such as exponential or trigonometric functions. Additionally, it may not work for more complex equations with multiple non-homogeneous terms.

## 4. Can the undetermined coefficients method be used for higher-order differential equations?

Yes, the undetermined coefficients method can be used for higher-order differential equations. However, the process becomes more complex as the number of undetermined coefficients increases. It is also important to note that the method is most effective for second-order differential equations.

## 5. Are there other methods for solving initial value problems?

Yes, there are other methods for solving initial value problems, such as the variation of parameters method and the Laplace transform method. These methods may be more suitable for certain types of equations and can be used as alternatives to the undetermined coefficients method.

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