# Set up appropriate form of solution Yp.

• Jul 25th 2009, 12:30 PM
DMDil
Set up appropriate form of solution Yp.
Set up the appropriate form of a solution yp of y^(3)+y'' = 4x^2 , but DO NOT determine the values of the coefficients.

guy this is what i have done so far

DE : y^(3)+y'' = 4x^2
CE: r^3 + r'' = 0
r1 = 0, r2= 0 , r3= -1

yc = C1 + C2x + C3 e^-x

now here i am confuse about what to do to guess yp. should yp be Ax^2?? plz help me understand this (Itwasntme)(Crying)
• Jul 25th 2009, 01:29 PM
Jester
Quote:

Originally Posted by DMDil
Set up the appropriate form of a solution yp of y^(3)+y'' = 4x^2 , but DO NOT determine the values of the coefficients.

guy this is what i have done so far

DE : y^(3)+y'' = 4x^2
CE: r^3 + r'' = 0
r1 = 0, r2= 0 , r3= -1

yc = C1 + C2x + C3 e^-x

now here i am confuse about what to do to guess yp. should yp be Ax^2?? plz help me understand this (Itwasntme)(Crying)

Since $y = 1$ and $y = x$ are part of your complimentary solution, then you'll need to seek a particular solution of the form

$
y_p = A x^4 + B x^3 + C x^2
$
• Jul 25th 2009, 02:13 PM
DMDil
how u came up with Ax^4+Bx^3,, http://www.mathhelpforum.com/math-he...97834e03-1.gif i meant how do i decided Yp in every equation?
• Jul 25th 2009, 02:30 PM
Jester
Quote:

Originally Posted by DMDil
how u came up with Ax^4+Bx^3,, http://www.mathhelpforum.com/math-he...97834e03-1.gif i meant how do i decided Yp in every equation?

If the rhs is some power of x then you would need to seek a particluar solution of the form

$
y = c_0 + c_1 x + c_2 x^2 + \cdots + c_n x^n
$

depending on the power $n$. Now if any of your complimentary solution contains any of these terms then you need to raise the power by the number of terms that the complimentary solution contains. In your case it contains $1$ and $x$ so you need to "bump" up the particular solutions by 2.
• Jul 25th 2009, 05:19 PM
DMDil
i am not sure if it is allow to ask same problem again with different values(Itwasntme), what would Yp be if i have y^3 +9y' = x^2 exp(x)
for Yc, i would ge Yc = C1+C2cos(3x)+C3sin(3x)
how would i guess Yp now...is there any kind of table which can guide.
• Jul 25th 2009, 05:37 PM
defog171
This is third order linear non-homogeneous ODE which you have to seperate answers which is the $Y_p$ and $Y_h$ which the final solution would be the $Y(x) = Y_p + Y_h$ , and $Y_p$ is the solution to the left hand side of the ODE and $Y_h$ is the solution to the right hand side of the ODE. $Y_h$ can be solved using the Laplace transformation.
• Jul 25th 2009, 06:08 PM
DMDil
i have not learn Laplace transformation yet. but i do know in order to solve this problem complete i have to use y(x) = Yc +Yp, Yc is easy to find since Yc is same in homogenous. but guess Yp is new thing for me. in every problem is different rule. what is the basic thing to follow here. what is it that i need to do to figure out Yp correctly?
• Jul 25th 2009, 06:49 PM
defog171
Quote:

Originally Posted by defog171
This is third order linear non-homogeneous ODE which you have to seperate answers which is the $Y_p$ and $Y_h$ which the final solution would be the $Y(x) = Y_p + Y_h$ , and $Y_p$ is the solution to the left hand side of the ODE and $Y_h$ is the solution to the right hand side of the ODE. $Y_h$ can be solved using the Laplace transformation.

oops.. i did it again, sorry for the wrong info
should be like this
$Y(x) = Y_h + Y_p$ ,
$Y_h$ is the solution to the left hand side of the ODE which is also same as your $Y_c$
and $Y_p$ is the solution to the right hand side of the ODE that makes your ODE non-homogeneous.

to solve $Y_p$, you need to learn the (method of undetermined coefficients) or variation of parameters (if you have the basis of the solutions) which you need to find the Wronskian. But for this $y'''+9y' = x^2 e^x
$
i suggest you to learn (method of undetermined coefficients) first.
• Jul 25th 2009, 07:41 PM
defog171
$y_p=C(x^2e^x)$
thus,
$y_p'=C(x^2e^{x}+2xe^{x})$
$y_p''=C(x^2e^{x}+2xe^{x}+2xe^{x}+2e^{x})=C(x^2e^{x }+4xe^{x}+2e^{x})$
$y_p'''=C(x^2e^x+2xe^{x}+4xe^{x}+4e^{x}+2e^{x})=C(x ^2e^{x}+6xe^{x}+6e^{x})$

substitute this $y_p$ and it's derivatives into the ODE, gives

$C(x^2e^x+6xe^x+6e^x) +9C(x^2e^x+2xe^x) = x^2e^x$

rearranging the equation yields

$e^x(10Cx^2+6Cx+24C) = e^x(x^2)$

by comparing coefficients for the both sides, for $x^2,$ $10C = 1,$ therefore $C=1/10,$

then $y_p = \frac{1}{10}x^2e^x$

can anybody verify this.. i think it's wrong
maybe if we choose $y_p=C_1x^2e^x+C_2xe^x+C_3e^x$
it will be better?
• Jul 27th 2009, 04:32 AM
DMDil
i think i only have to solve up to Yp = A(x^2 exp(x)), since the problem says Do not Determine the value of the coefficent