Finding particular solution to differential equations

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SUMMARY

The forum discussion focuses on finding particular solutions to non-homogeneous ordinary differential equations (ODEs). The two equations discussed are y''' - 9y'' + 14y' = x² and y'' - 9y' + 14y = x²e^(4x). The initial guesses for the particular solutions, yp = Px³ + Qx² + Rx and yp = (Px² + Qx + R)e^(4x), were incorrect. The correct approach involves recognizing the need for a general solution of the form y = A(x²e^(4x)) for the second equation and ensuring that all derivatives satisfy the original equation.

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lillybeans
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There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x

For the first one I wrote:
yp=Px3+Qx2+Rx

Second one:
yp=(Px2+Qx+R)e4x

Neither of the answers are correct, according to the computer. Where did I go wrong?

Thanks.
 
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The answers are not correct because you're not solving the equations correctly... First do you understand that your equations are non-homogenous?
 
Zondrina said:
The answers are not correct because you're not solving the equations correctly... First do you understand that your equations are non-homogenous?

Yes, of course, that's why I am asked to find a particular solution. Where did I go wrong? please let me know!
 
Okay, so take for example your second question. Notice that your derivatives and original function have to add up to x2e4x?

So your general solution must have the form : y = A(x2e4x) where A is some constant number.

Now take the required derivatives and see if they satisfy your equation. ( You'll get a bunch of constants which you have to add together and solve for A for the particular solution you want ).
 
Zondrina said:
Okay, so take for example your second question. Notice that your derivatives and original function have to add up to x2e4x?

So your general solution must have the form : y = A(x2e4x) where A is some constant number.

Now take the required derivatives and see if they satisfy your equation. ( You'll get a bunch of constants which you have to add together and solve for A for the particular solution you want ).

I don't quite understand why your guess for the particular solution has the form Ax2e4x as opposed to (Ax2+Bx+C)e4x. Shouldn't you use the general form of the second degree polynomial?
 
A non-homogenous equation with constant coefficients has the general solution y = Ag(t) where A is a constant.
 
Zondrina said:
A non-homogenous equation with constant coefficients has the general solution y = Ag(t) where A is a constant.

Then suppose for y''-9y'+14y my yp=Ax2e4x

Then

y'=2Axe4x+4Ax2e4x
y''=2Ae4x+8Axe4x+8Axe4x+16Ax2e4x

When you plug this all back into the original equation, equate the coefficients to solve for A, it doesn't work. Need more than just one constant.
 

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