MHB Jerome's questions at Yahoo Answers regarding inhomogeneous linear ODEs

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Linear Odes
AI Thread Summary
The discussion focuses on solving two inhomogeneous linear ordinary differential equations (ODEs) using the method of undetermined coefficients. The first ODE is y''' + y' = 4sin(x) + 2x^2, where the general solution combines a homogeneous solution and a particular solution derived from the annihilator method. The second ODE, y''' + y'' + 3y' - 5y = 5sin(2x) + 10x^2 - 3x + 7, follows a similar approach, with the particular solution determined through coefficient comparison after substituting derivatives into the equation. Both solutions illustrate the application of the method of undetermined coefficients and the annihilator technique to find particular solutions. The thread encourages further discussion and problem-solving on differential equations.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
Hello jerome,

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

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

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

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

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

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

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

The characteristic roots are then:

$$r=0,\,\pm i$$

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

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

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

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

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

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

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 $$y_p'(x)$$ and $$y_p'''(x)$$, substitute these into (1), and solve the resulting linear system that arises when we equate coefficients. So, computing the necessary derivatives, we find:

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

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

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

Substituting into (1), we find:

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

Rearranging, we may write:

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

Equating coefficients, we find:

$$c_6=0$$

$$c_8=-2$$

$$c_4=\frac{2}{3}$$

$$c_3=0$$

$$c_2=-4$$

Hence, we find:

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

and by the principle of superposition, we find:

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

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

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

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

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

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

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

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

The characteristic roots are then:

$$r=0,\pm2i,1,-1\pm2i$$

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

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

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

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

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

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

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 $$y_p'(x),\,y_p''(x),\,y_p..'(x)$$ and substitute these into (1), and solve the resulting linear system that arises when we equate coefficients. So, computing the necessary derivatives, we find:

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

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

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

Substituting into (1), we find:

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

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

Rearranging, we may write:

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

Equating coefficients, we find:

$$-9c_4-2c_5=0$$

$$2c_4-9c_5=5$$

$$-5c_3=10$$

$$-5c_2+6c_3=-3$$

$$-5c_1+3c_2+2c_3=7$$

Solving this system, we find:

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

Hence, we find:

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

and by the principle of superposition, we find:

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

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.
 
Last edited:
the method i understand is the undetermined coefficients, i don't understand all these sir
 
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.) $$y'''+y'=4\sin(x)+2x^2$$

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

$$y'''+y'=0$$

The characteristic or auxiliary equation is:

$$r^3+r=r(r^2+1)=0$$

and so the characteristic roots are:

$$r=0,\,\pm i$$

and so we find:

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

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

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

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

$$4\sin(x)$$

which is of the form:

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

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

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

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:

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

Now, for the term:

$$2x^2$$

which is of the form:

$$a_2x^2+a_1x+a_0$$

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

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

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

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

By superposition, we then have:

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

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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top