Method of undetermined coefficients

• MHB
• ineedhelpnow
In summary, the method of undetermined coefficients was used to find the general solution of the ODE $y''+3y''+3y'+y=e^{x}-x-1$. By constructing a particular solution based on the RHS, and equating coefficients, the particular solution was found to be $y_p(x)=\frac{1}{8}e^x-x+2$, and the general solution was determined to be $y(x)=e^{-x}\left(c_3x^2+c_2x+c_1\right)+\frac{1}{8}e^x-x+2$.
ineedhelpnow
Use the method of undetermined coefficients to find a general solution of the ODE:
$y''+3y'+2y=2x^{2}+4x+5$$r^{2}+3r+2=0$
$r=-2$ and $r=-1$
$y_{h}=c_{1}e^{-2x}+c_{2}e^{-x}$
I'm not sure how to get $y_{p}$ here

So here's what I've done so far. I have my final exam tomorrow and I have a few questions to go over as a review. I need help with the rest of this!

Here's another one:
$y''+3y''+3y'+y=e^{x}-x-1$

$r^{3}+3r^{2}+3r+1=0$
$r_{1,2,3}=-1$
$y_{h}=c_{1}e^{-x}+xe^{-x}(c_{2}+c_{3})$
Not really sure how to get $y_{p}$ here either

1.) You have correctly identified the homogeneous solution, and so if we consult a table, we find that the particular solution must take the form:

$$\displaystyle y_p(x)=Ax^2+Bx+C$$

Then compute the needed derivatives, and substitute into the original ODE, and the equate the corresponding coefficients to determine $A,\,B,\,C$.

Can you proceed?

2.) You have correctly identified that the characteristic equation has the root $r=-1$ of multiplicity 3, and so the homogeneous solution will take the form:

$$\displaystyle y_h(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}$$

Consulting a table, and applying the principle of superposition, we find the particular solution must take the form:

$$\displaystyle y_p(x)=Ae^x+Bx+C$$

Can you proceed? :)

I understand. But can you please post the table or a picture of it?! Because I can't find it anywhere in my book but I know there is one because I have referenced it before.

Here is the table (along with its justification):

http://mathhelpboards.com/math-notes-49/justifying-method-undetermined-coefficients-4839.html

I don't really understand what Pn is. :(

ineedhelpnow said:
I don't really understand what Pn is. :(

If $g(x)$ contains the $n$th degree polynomial $p_n(x)$, then the particular solution (in types I and IV) will contain the $n$th degree polynomial $P_n(x)$.

the particular solution is chosen based on the RHS, correct?

ineedhelpnow said:
the particular solution is chosen based on the RHS, correct?

Yes, if the ODE is of the form:

$$\displaystyle L[y](x)=g(x)$$

then we base the form of our particular solution on $g(x)$. If you prefer not to use a table, you can use the annihilator method to find the form of the particular solution:

I figured the first one out but I still can't get around the second.

Okay, the second one is:

$$\displaystyle y'''+3y''+3y'+y=e^{x}-x-1$$

We have already established the homogeneous solution is:

$$\displaystyle y_h(x)=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}=e^{-x}\left(c_3x^2+c_2x+c_1\right)$$

Next, we posit that the particular solution must take the form:

$$\displaystyle y_p(x)=Ae^x+Bx+C$$

So, we need to compute:

$$\displaystyle y_p'(x)=Ae^x+B$$

$$\displaystyle y_p''(x)=Ae^x$$

$$\displaystyle y_p'''(x)=Ae^x$$

Now, substituting these into the given ODE, there results:

$$\displaystyle Ae^x+3Ae^x+3(Ae^x+B)+Ae^x+Bx+C=e^{x}-x-1$$

$$\displaystyle 8Ae^x+Bx+3B+C=e^{x}-x-1$$

By equating corresponding coefficients, we obtain:

$$\displaystyle 8A=1\implies A=\frac{1}{8}$$

$$\displaystyle B=-1$$

$$\displaystyle 3B+C=-1\implies C=2$$

Thus, the particular solution is:

$$\displaystyle y_p(x)=\frac{1}{8}e^x-x+2$$

And so the general solution is:

$$\displaystyle y(x)=y_h(x)+y_p(x)=e^{-x}\left(c_3x^2+c_2x+c_1\right)+\frac{1}{8}e^x-x+2$$

What is the Method of Undetermined Coefficients?

The Method of Undetermined Coefficients is a technique used in mathematics, specifically in differential equations, to solve for a particular solution that satisfies a given non-homogeneous equation. It is used to find the coefficients of a linear combination of known functions that will satisfy the equation.

When is the Method of Undetermined Coefficients used?

This method is used when solving non-homogeneous equations, where the right-hand side of the equation is a function that cannot be expressed as a linear combination of the functions on the left-hand side.

How does the Method of Undetermined Coefficients work?

The method involves first finding the complementary solution, which is the solution to the homogeneous equation. Then, a particular solution is assumed to be a linear combination of known functions, such as polynomials, exponential functions, or trigonometric functions. The coefficients of these functions are then determined by plugging them into the original equation and solving for the unknown coefficients.

What are the limitations of the Method of Undetermined Coefficients?

The method is only applicable to linear, constant coefficient differential equations. It also requires that the non-homogeneous term is a simple function, and that the complementary solution and particular solution are linearly independent.

Are there any other methods for solving non-homogeneous equations?

Yes, there are other methods such as the Variation of Parameters method and the Laplace Transform method. These methods may be more suitable for certain types of non-homogeneous equations, so it is important to understand all available methods and use the most appropriate one for a given equation.

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