Non-homogeneous Equations problem 1

Here is the problem:
y''-10 y' +25y = -16 e^(5t)

And I am trying to find the particular solution!

I tried using the method of undetermined coefficients but no matter what my initial guess of Yp(x) I always get 0 on the left side

For example:

Yp(x) = Axe^(5x)
Y'p(x) = A(5x+1)e^(5x)
y''p(x) = A(25x+10)e^(5x)

Then I substitute into the original function

and I get:

(25AX-50AX+25AX+10A-10A)e^(5x) = -16 e^(5x)
-> 0(e^(5x) = -16 e^(5x)

then I don't know what to do anymore...
Did I do something wrong or I should use a different method to approach this problem?

Thank you for the help!!

Writing the DE equation as...

$\displaystyle y^{''} -10\cdot y^{'} + 25\cdot y = g(t)$ (1)

... the general solution of the homogeneus equation [i.e. the equation in which is $\displaystyle g(x)=0$...] is...

$\displaystyle y(t)= c_{1}\cdot e^{5t} + c_{2}\cdot t\cdot e^{5t}$ (2)

In the case You have proposed is $\displaystyle g(t)= -16\cdot e^{5t}$, so that $\displaystyle g(t)$ is solution of the homogeneous equation and that can produce some problems. In almost all such cases the solution of the problem requires the use of Laplace Transform. Writing (1) in term of LT we obtain...

$\displaystyle s^{2}\cdot Y(s) -10\cdot s\cdot Y(s) + 25\cdot Y(s)= -\frac{16}{s-5}$ (3)

... so that is...

$\displaystyle Y(s)=-\frac{16}{(s-5)^{3}}$ (4)

The particular solution You are searching is the inverse LT of (4) and loocking at the 'manual' we find that this 'particular solution' $\displaystyle y^{*} (t)$ is [excluding errors of me (Thinking)] ...

$\displaystyle y^{*} (t) = -8 \cdot t^{2} e^{5t}$ (5)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Thank you so much for the help!!