correct method to solve differencial Considering an equation:
$$ 4y''-12y'+9y= e^{5t}+e^{3t}$$
Solving the homogeneous equation
$$ y_h= c_1e^{1.5t}+c_2te^{1.5t}$$
Now to find the general solution of the given equation, Can you resolve yourself by the following method?
$$ 4y_1''-12y_1'+9y_1= e^{5t}$$ and 
$$4y_2''-12y_2'+9y_2= e^{3t}$$
Then $$y_1 =Ae^{5t}$$ and $$y_2 =Be^{3t}$$
Making the first and second derivative of y1 e y2.
and substituting in the initial equation.
$$A=\frac{1}{49}$$
 and $$ B = \frac{1}{9}$$
Then the solution geral of equation is:
$$ y(t) =c_1e^{1.5t}+c_2te^{1.5t}+\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$
Is this a correct method to solve?
Thanks!!
 A: Yes, your solution is correct. The given equation is a second order, linear, nonhomogeneous differential equation with constant coefficients. The solution is given by
$$y_g(t)=y_h(t)+y_p(t)$$
where $y_g(t)$ is the general solution, $y_h(t)$ is the homogeneous solution, and $y_p(t)$ is the particular solution. You first found the homogeneous solution as 
$$y_h(t)= c_1e^{1.5t}+c_2te^{1.5t}$$
then to find the particular solution you need to solve
$$ 4y''-12y'+9y= e^{5t}+e^{3t}$$
from which you applied the superposition principle and the method of undetermined coefficients. To find the particular solution, you need to solve
$$4y''-12y'+9y= e^{5t}$$
and
$$4y''-12y'+9y= e^{3t}$$
and then use the superposition principle to add these two solutions together. The particular solution will therefore take the form
$$y_p(t)=y_{p_1}(t)+y_{p_2}(t)$$
The method of undetermined coefficients forms two guesses
$$y_{p_1}(t)=Ae^{5t},\quad y_{p_2}(t)=Be^{3t}$$
from which you can relate the coefficients to find
$$A=\frac{1}{49},\quad B=\frac{1}{9}$$
which implies that the particular solution is
$$y_p(t)=y_{p_1}(t)+y_{p_2}(t)=Ae^{5t}+Be^{3t}=\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$
therefore the general solution is given as
$$y_g(t)=y_h(t)+y_p(t)=c_1e^{1.5t}+c_2te^{1.5t}+\frac{1}{49}e^{5t}+\frac{1}{9}e^{3t}$$
