How do I "guess" the solution to y to this D.E.

y"-4y'+3y = 8xe^{x}

I guessed that the solution to y was Axe^{x} but when I plugged it into y it did not get me the correct answer. Please don't work out the problem for me, right now I just need help understanding how I would guess y in this situation. Much appreciated.

Re: How do I "guess" the solution to y to this D.E.

Quote:

Originally Posted by

**fb280** y"-4y'+3y = 8xe^{x}

I guessed that the solution to y was Axe^{x} but when I plugged it into y it did not get me the correct answer. Please don't work out the problem for me, right now I just need help understanding how I would guess y in this situation. Much appreciated.

Since the complimentary solution is $\displaystyle y_c=c_1e^{x}+c_2e^{3x}$

You would normally gues the solution is

$\displaystyle y=Ae^{x}+Bxe^{x}$

But since the first term is part of the complimentary solution we need to multiply the particular soluiton by $\displaystyle x$ to get the guess

$\displaystyle y_p=Axe^{x}+Bx^2e^{x}$

Re: How do I "guess" the solution to y to this D.E.

why would the guess normally be Ae^{x} for one part instead of Axe^{x}? And why is there a B? If there was a B I would have thought that it would be Be^{x} and not Bxe^{x}

Re: How do I "guess" the solution to y to this D.E.

On the right side of the ODE you have the product of a linear function and an exponential function, so you would expect the particular solution would take the form:

$\displaystyle y_p=(Ax+B)e^x$

Note: while I have switched A and B from the form **The EmptySet** used, they are equivalent.

As stated, you don't want any of the terms of your particular solution to be part of the complementary solution, so you multiply the particular solution by $\displaystyle x^s$ where $\displaystyle s$ is the smallest integer so that no terms of the particular solution are part of the complementary solution. In this case, $\displaystyle s=1$. Hence:

$\displaystyle y_p=x(Ax+B)e^x=Ax^2e^x+Bxe^x$

Re: How do I "guess" the solution to y to this D.E.

Okay I think I understand what I was confused on before. But when I plug in Ax^{2}e^{x}+Bxe^{x} I got stuck at the part where it turns out to be

-2A2xe^{x}+A2e^{x}+Bx4e^{x}-4Bx2e^{x}+3Bxe^{x}=8xe^{x}

Ax^{2}e^{x}-4Ax^{2}e^{x}+3Ax^{2}e^{x} cancels out so I did not include this part above

Re: How do I "guess" the solution to y to this D.E.

Using:

$\displaystyle y_p=Ax^2e^x+Bxe^x$

we find:

$\displaystyle y_p'=e^x(Ax^2+2Ax+Bx+B)$

$\displaystyle y_p''=e^x(Ax^2+4Ax+Bx+2A+2B)$

and substituting into the ODE, we get:

$\displaystyle e^x(Ax^2+4Ax+Bx+2A+2B)-4e^x(Ax^2+2Ax+Bx+B)+3e^x(Ax^2+Bx)=8xe^x$

Divide though by $\displaystyle e^x$, distribute constants and collect like terms:

$\displaystyle Ax^2+4Ax+Bx+2A+2B-4(Ax^2+2Ax+Bx+B)+3(Ax^2+Bx)=8x$

$\displaystyle Ax^2+4Ax+Bx+2A+2B-4Ax^2-8Ax-4Bx-4B+3Ax^2+3Bx=8x$

$\displaystyle (-4A)x+(2A-2B)=8x+0$

Equating coefficients yields the system:

$\displaystyle -4A=8$

$\displaystyle 2A-2B=0$

Now, solve this system to determine *A* and *B*, and then you have your particular solution.

Re: How do I "guess" the solution to y to this D.E.

okay I understand where I went wrong now. thanks for replying, this really helped me a lot.