Differential Equation Initial Value Problem

  • #1

Homework Statement



I managed to work this problem all the way through, but I am in no way certain of my answer. I'd greatly appreciate any insight!

Find the solution of the initial value problem.

y'''+4y'=x, y(0)=y'(0)=0, y''(0)=1

Homework Equations



Just for clarification purposes, my professor uses yc for the general homogeneous solution; I've seen yh used and didn't want to cause confusion.

fn(x)y(n)+fn-1(x)y(n-1)+...+f1(x)y'+f0(x)y=0

General solution: yc=c1y1(x)+c2y2(x)+...+cnyn(x)

The Attempt at a Solution



Note: m is used as the variable for characteristic equations rather than r, which I've also seen.

First, to find yc, I set the left side of the equation equal to zero and found the characteristic equation m3+4m=0, from which I found roots 0, 2i, -2i. I converted them into real solutions and ended up with yc=c1+c2cos(2x)+c3sin(2x).

To find c1 and c2, I took the first and second derivatives of yc and plugged in the initial values as given. I ended up with yc=(1/4)-(1/4)cos(2x).

Next, to find yp, I guessed yp=Ax3+Bx2+Cx+D and plugged its first and third derivatives into the original differential equation. I ended up finding A=0, B=1/8, C=-1/2. That gave me yp=(1/8)x2-(1/2)x.

I got the general solution to the heterogeneous differential equation y=yc+yp:
y=(1/4)-(1/4)cos(2x)+(1/8)x2-(1/2)x.
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



I managed to work this problem all the way through, but I am in no way certain of my answer. I'd greatly appreciate any insight!

Find the solution of the initial value problem.

y'''+4y'=x, y(0)=y'(0)=0, y''(0)=1

Homework Equations



Just for clarification purposes, my professor uses yc for the general homogeneous solution; I've seen yh used and didn't want to cause confusion.

fn(x)y(n)+fn-1(x)y(n-1)+...+f1(x)y'+f0(x)y=0

General solution: yc=c1y1(x)+c2y2(x)+...+cnyn(x)

The Attempt at a Solution



Note: m is used as the variable for characteristic equations rather than r, which I've also seen.

First, to find yc, I set the left side of the equation equal to zero and found the characteristic equation m3+4m=0, from which I found roots 0, 2i, -2i. I converted them into real solutions and ended up with yc=c1+c2cos(2x)+c3sin(2x).

To find c1 and c2, I took the first and second derivatives of yc and plugged in the initial values as given. I ended up with yc=(1/4)-(1/4)cos(2x).

Next, to find yp, I guessed yp=Ax3+Bx2+Cx+D and plugged its first and third derivatives into the original differential equation. I ended up finding A=0, B=1/8, C=-1/2. That gave me yp=(1/8)x2-(1/2)x.

I got the general solution to the heterogeneous differential equation y=yc+yp:
y=(1/4)-(1/4)cos(2x)+(1/8)x2-(1/2)x.

I didn't check your work, but I can tell you that you must first write the general solution ##y = y_c + y_p## before you evaluate the constants with the initial conditions.
 
  • #3
I thought that might be the case, but I tried it and came up with a bunch of long, crazy rational expressions with several radicals. I'll try again.

If anyone reading knows how to do it correctly, I'd be eternally grateful for your help!
 
  • #4
LCKurtz
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Your ##y_p## isn't correct. You should get ##y_p = \frac{x^2} 8##. You can check that it works. And you won't get radicals anywhere.
 
  • #5
I'll start working on getting it right now; thank you so much!
 

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