Jerome's questions at Yahoo Answers regarding inhomogeneous linear ODEs

[MarkFL]

Here are the questions:

HELP WITH THESE DIFFERENTIAL EQUATION?

please use the method of undetermined coefficients and show your steps with explanations

(1) Y"'+y'=4sinx +2x^2
(2) Y'''+y"+3y'-5y=5sin2x+10x^2-3x+7

Here is a link to the questions:

HELP WITH THESE DIFFERENTIAL EQUATION? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello jerome,

1.) We are given to solve the inhomogeneous linear 3rd order ODE:

(1) \(\displaystyle y'''+y'=4\sin(x)+2x^2\)

The differential operator \(\displaystyle D^2+1\) annihilates \(\displaystyle 4\sin(x)\) and \(\displaystyle D^3\) annihilates \(\displaystyle 2x^2\) hence the operator:

\(\displaystyle A\equiv D^3(D^2+1)\)

annihilates \(\displaystyle 4\sin(x)+2x^2\).

Thus, applying \(\displaystyle A\) to both sides of (1) gives us:

(2) \(\displaystyle D^4(D^2+1)^2[y]=0\)

The characteristic roots are then:

\(\displaystyle r=0,\,\pm i\)

where \(\displaystyle r=0\) is of multiplicity 4 and \(\displaystyle r=\pm i\) are of multiplicity 2, and so the general solution to (2) is given by:

(3) \(\displaystyle y(x)=c_1+c_2x+c_3x^2+c_4x^3+(c_5+c_6x)\cos(x)+(c_7+c_8x)\sin(x)\)

Now, recall that a general solution to (1) is of the form \(\displaystyle y(x)=y_h(x)+y_p(x)\). Since every solution to (1) is also a solution to (2), then \(\displaystyle y(x)\) must have the form displayed on the right-hand side of (3). But we recognize that:

\(\displaystyle y_h(x)=c_1+c_2\cos(x)+c_3\sin(x)\)

and so there must exist a particular solution of the form:

\(\displaystyle y_p(x)=c_2x+c_3x^2+c_4x^3+c_6x\cos(x)+c_8x\sin(x)\)

Now it is a matter of using the method of undetermined coefficients to determine the particular solution that satisfies (1).

So, what we need to do is compute \(\displaystyle y_p'(x)\) and \(\displaystyle y_p'''(x)\), substitute these into (1), and solve the resulting linear system that arises when we equate coefficients. So, computing the necessary derivatives, we find:

\(\displaystyle y_p'(x)=c_2+2c_3x+3c_4x^2+c_6(\cos(x)-x\sin(x))+c_8(\sin(x)+x\cos(x))\)

\(\displaystyle y_p''(x)=2c_3+6c_4x+c_6(-2\sin(x)-x\cos(x))+c_8(2\cos(x)-x\sin(x))\)

\(\displaystyle y_p'''(x)=6c_4+c_6(-3\cos(x)+x\sin(x))+c_8(-3\sin(x)-x\cos(x))\)

Substituting into (1), we find:

\(\displaystyle \left(6c_4+c_6(-3\cos(x)+x\sin(x))+c_8(-3\sin(x)-x\cos(x)) \right)+\left(c_2+2c_3x+3c_4x^2+c_6(\cos(x)-x\sin(x))+c_8(\sin(x)+x\cos(x)) \right)=4\sin(x)+2x^2\)

Rearranging, we may write:

\(\displaystyle -2c_6\cos(x)-2c_8\sin(x)+3c_4x^2+2c_3x+c_2+6c_4=0\cos(x)+4\sin(x)+2x^2+0x+0\)

Equating coefficients, we find:

\(\displaystyle c_6=0\)

\(\displaystyle c_8=-2\)

\(\displaystyle c_4=\frac{2}{3}\)

\(\displaystyle c_3=0\)

\(\displaystyle c_2=-4\)

Hence, we find:

\(\displaystyle y_p(x)=-4x+\frac{2}{3}x^3-2x\sin(x)\)

and by the principle of superposition, we find:

\(\displaystyle y(x)=c_1+c_2\cos(x)+c_3\sin(x)-4x+\frac{2}{3}x^3-2x\sin(x)\)

2.) We are given to solve the inhomogeneous linear 3rd order ODE:

(1) \(\displaystyle y'''+y''+3y'-5y=5\sin(2x)+10x^2-3x+7\)

The differential operator \(\displaystyle \frac{D^2}{2^2}+1\) annihilates \(\displaystyle 5\sin(2x)\) and \(\displaystyle D^3\) annihilates \(\displaystyle 10x^2-3x+7\) hence the operator:

\(\displaystyle A\equiv \frac{1}{4}D^3(D^2+4)\)

annihilates \(\displaystyle 5\sin(2x)+10x^2-3x+7\).

Thus, applying \(\displaystyle A\) to both sides of (1) gives us:

(2) \(\displaystyle D^3(D^2+4)(D-1)(D^2+2D+5)[y]=0\)

The characteristic roots are then:

\(\displaystyle r=0,\pm2i,1,-1\pm2i\)

The root \(\displaystyle r=0\) is of multiplicity 3, and so the general solution to (2) is given by:

\(\displaystyle y(x)=c_1+c_2x+c_3x^2+c_4\cos(2x)+c_5\sin(2x)+c_6e^x+c_7e^{-x}\cos(2x)+c_8e^{-x}\sin(2x)\)

Now, recall that a general solution to (1) is of the form \(\displaystyle y(x)=y_h(x)+y_p(x)\). Since every solution to (1) is also a solution to (2), then \(\displaystyle y(x)\) must have the form displayed on the right-hand side of (3). But we recognize that:

\(\displaystyle y_h(x)=c_6e^x+c_7e^{-x}\cos(2x)+c_8e^{-x}\sin(2x)\)

and so there must exist a particular solution of the form:

\(\displaystyle y_p(x)=c_1+c_2x+c_3x^2+c_4\cos(2x)+c_5\sin(2x)\)

Now it is a matter of using the method of undetermined coefficients to determine the particular solution that satisfies (1).

So, what we need to do is compute \(\displaystyle y_p'(x),\,y_p''(x),\,y_p..'(x)\) and substitute these into (1), and solve the resulting linear system that arises when we equate coefficients. So, computing the necessary derivatives, we find:

\(\displaystyle y_p'(x)=c_2+2c_3x-2c_4\sin(2x)+2c_5\cos(2x)\)

\(\displaystyle y_p''(x)=2c_3-4c_4\cos(2x)-4c_5\sin(2x)\)

\(\displaystyle y_p'''(x)=8c_4\sin(2x)-8c_5\cos(2x)\)

Substituting into (1), we find:

\(\displaystyle \left(8c_4\sin(2x)-8c_5\cos(2x) \right)+\left(2c_3-4c_4\cos(2x)-4c_5\sin(2x) \right)+3\left(c_2+2c_3x-2c_4\sin(2x)+2c_5\cos(2x) \right)-\)

\(\displaystyle \,\,\,\,\,\,\,\,5\left(c_1+c_2x+c_3x^2+c_4\cos(2x)+c_5\sin(2x) \right)=5\sin(2x)+10x^2-3x+7\)

Rearranging, we may write:

\(\displaystyle (-9c_4-2c_5)\cos(2x)+(2c_4-9c_5)\sin(2x)+(-5c_3)x^2+(-5c_2+6c_3)x+(-5c_1+3c_2+2c_3)=0\cos(2x)+5\sin(2x)+10x^2-3x+7\)

Equating coefficients, we find:

\(\displaystyle -9c_4-2c_5=0\)

\(\displaystyle 2c_4-9c_5=5\)

\(\displaystyle -5c_3=10\)

\(\displaystyle -5c_2+6c_3=-3\)

\(\displaystyle -5c_1+3c_2+2c_3=7\)

Solving this system, we find:

\(\displaystyle c_1=-\frac{82}{25},\,c_2=-\frac{9}{5},\,c_3=-2,\,c_4=\frac{2}{17},\,c_5=-\frac{9}{17}\)

Hence, we find:

\(\displaystyle y_p(x)=-\frac{82}{25}-\frac{9}{5}x-2x^2+\frac{2}{17}\cos(2x)-\frac{9}{17}\sin(2x)\)

and by the principle of superposition, we find:

\(\displaystyle y(x)=c_1e^x+c_2e^{-x}\cos(2x)+c_3e^{-x}\sin(2x)-\frac{82}{25}-\frac{9}{5}x-2x^2+\frac{2}{17}\cos(2x)-\frac{9}{17}\sin(2x)\)

To jerome and any other guests viewing this topic, I invite and encourage you to post other differential equations problems in our http://www.mathhelpboards.com/f17/ forum.

Best Regards,

Mark.

 

[Jerome1]

the method i understand is the undetermined coefficients, i don't understand all these sir

 

[MarkFL]

Rather than use a look-up table, I used the annihilator method to determine the form of the particular solution, and then I used the method of undetermined coefficients to determine the actual particular solution. I will demonstrate how to determine the form of the particular solution for the first problem using a table, and then I want to see if you can apply this to the second problem.

1.) \(\displaystyle y'''+y'=4\sin(x)+2x^2\)

First, we want to find the corresponding homogeneous solution \(\displaystyle y_h(x)\), i.e., the solution to:

\(\displaystyle y'''+y'=0\)

The characteristic or auxiliary equation is:

\(\displaystyle r^3+r=r(r^2+1)=0\)

and so the characteristic roots are:

\(\displaystyle r=0,\,\pm i\)

and so we find:

\(\displaystyle y_h(x)=c_1+c_2\cos(x)+c_2\sin(x)\)

Now, we look at the right-hand side of the original ODE, which is:

\(\displaystyle 4\sin(x)+2x^2\)

Now, referring to a table, we find for the term:

\(\displaystyle 4\sin(x)\)

which is of the form:

\(\displaystyle a\cos(\beta x)+b\sin(\beta x)\)

that the particular solution associated with this term must be of the form:

\(\displaystyle y_{p_1}(x)=x^s\left(A\cos(\beta x)+B\sin(\beta x) \right)\)

where the non-negative integer $s$ is chosen to be the smallest integer so that no term in the particular solution is a solution to the corresponding homogeneous equation. Thus, as you can see, we find in this case we require $s=1$, and so:

\(\displaystyle y_{p_1}(x)=x\left(A\cos(\beta x)+B\sin(\beta x) \right)\)

Now, for the term:

\(\displaystyle 2x^2\)

which is of the form:

\(\displaystyle a_2x^2+a_1x+a_0\)

we find that the particular solution associated with this term must be of the form:

\(\displaystyle y_{p_2}(x)=x^s\left(Cx^2+Dx+E \right)\)

Since \(\displaystyle y_h(x)\) has a constant term, we find \(\displaystyle s=1\), and so:

\(\displaystyle y_{p_2}(x)=x\left(Cx^2+Dx+E \right)\)

By superposition, we then have:

\(\displaystyle y_p(x)=y_{p_1}(x)+y_{p_2}(x)=x\left(A\cos(\beta x)+B\sin(\beta x) \right)+x\left(Cx^2+Dx+E \right)\)

As you can see this is the same form I determined above by using the annihilator method.

Now, see if you can apply this technique to the second problem, and post what you find, and I will be more than happy to help if you get stuck. :D

 

[CellCoree]

Hi Jerome,

Thank you for your questions about inhomogeneous linear ODEs. I will be happy to provide some guidance and explanations for using the method of undetermined coefficients to solve these equations.

Before we dive into the specific problems you have posted, let's review the general steps for using the method of undetermined coefficients to solve an inhomogeneous linear ODE:

Step 1: Identify the homogeneous solution
The first step is to find the solution to the corresponding homogeneous equation, which is obtained by setting all the non-homogeneous terms (such as sinx or 10x^2) equal to zero. This solution will be in the form of y_h = c1y1 + c2y2 + ... + cnyn, where c1, c2, ..., cn are arbitrary constants and y1, y2, ..., yn are linearly independent solutions to the homogeneous equation.

Step 2: Guess the form of the particular solution
Next, we need to guess the form of the particular solution, which will depend on the form of the non-homogeneous terms in the equation. For example, if the non-homogeneous term is sinx, then the particular solution will be of the form y_p = Asinx + Bcosx. If the non-homogeneous term is 10x^2, then the particular solution will be of the form y_p = Ax^2 + Bx + C.

Step 3: Substitute the guess into the equation
Once we have guessed the form of the particular solution, we substitute it into the original equation and solve for the coefficients in our guess. This will give us a particular solution y_p.

Step 4: Combine the homogeneous and particular solutions
Finally, we combine the homogeneous and particular solutions to get the general solution to the inhomogeneous equation, which is given by y = y_h + y_p.

Now, let's apply these steps to your specific equations:

(1) Y"'+y'=4sinx +2x^2
For this equation, we first need to find the solution to the corresponding homogeneous equation, which is y_h = c1e^(-x) + c2. Next, we guess that the particular solution will be of the form y_p = Asinx + Bcosx + Cx^2 + Dx + E. Substituting this into the equation and solving for the coefficients, we get y_p = -4