Differential Equation Initial Value Problem

[heartilly89]

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 y_c for the general homogeneous solution; I've seen y_h used and didn't want to cause confusion.

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

General solution: y_c=c₁y₁(x)+c₂y₂(x)+...+c_ny_n(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 y_c, 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 y_c=c₁+c₂cos(2x)+c₃sin(2x).

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

Next, to find y_p, I guessed y_p=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 y_p=(1/8)x²-(1/2)x.

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

 

[LCKurtz]

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 y_c for the general homogeneous solution; I've seen y_h used and didn't want to cause confusion.

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

General solution: y_c=c₁y₁(x)+c₂y₂(x)+...+c_ny_n(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 y_c, 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 y_c=c₁+c₂cos(2x)+c₃sin(2x).

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

Next, to find y_p, I guessed y_p=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 y_p=(1/8)x²-(1/2)x.

I got the general solution to the heterogeneous differential equation y=y_c+y_p:
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.

 

[heartilly89]

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!

 

[LCKurtz]

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.

 

[heartilly89]

I'll start working on getting it right now; thank you so much!