Let $~y_{1}(t)$,$~y_{2}(t)$ be solutions of the diff. eq. $y ''+ e^{t}y '+y = t^{3} $ Which of the following is a solution of the ODE? Let $~y_{1}(t)$,$~y_{2}(t)$ be solutions of the diff. eq. $y ''+ e^{t}y '+y = t^{3} $ Which of the following is a solution of the ODE?


*

*$y_{1}(t)-y_{2}(t)$ 

*$y_{1}(t)+y_{2}(t)$

*$y_{1}(t)-5y_{2}(t)$ 

*$2y_{1}(t)-y_{2}(t)$ 

*$2y_{1}(t)+3y_{2}(t)$ 
I'm not sure how to approach this problem and was hoping someone would be willing to assist.
 A: We have $y_i''+e^ty_i'+y_i=t^3$ for $i=1,2$. Suppose $y(t)=my_1(t)+ny_2(t)$ is a solution of the ODE, $m,n$ constant. Plug this into the ODE to get:$$m(y_1''+e^ty_1'+y_1)+n(y_2''+e^ty_2'+y_2)=t^3$$This gives $(m+n)t^3=t^3$, that is $m+n=1$. The correct answer is...?
A: Let $L$ denote the linear differential operator
$L \equiv \dfrac{d^2}{dt} + e^t \dfrac{d}{dt} + 1, \tag 1$
so that the differential equation of interest may be written
$Ly = t^3; \tag 2$
thus $y_1$ and $y_2$ each satisfy (2):
$Ly_1 = t^3, \; Ly_2 = t^3; \tag 3$
then it follows from the linearity of $L$ that
$L(y_1 + y_2) = Ly_1 + Ly_2 = t^3 + t^3 = 2t^3, \tag 4$
and also that 
$L(y_1 - y_2) = Ly_1 - Ly_2 = t^3 - t^3 = 0; \tag 5$
thus neither $y_1 + y_2$ nor $y_1 - y_2$ are solutions of (1)-(2).  We may likewise rule out (3) and (5):
$L(y_1 - 5y_2) = Ly_1 - 5Ly_2 = t^3 - 5t^3 = -4t^3, \tag 6$
$L(2y_1 + 3y_2) = 2Ly_1 + 3Ly_2 = 2t^3 + 3t^3 = 5t^3. \tag 7$
As for (4),
$L(2y_1 - y_2) = 2Ly_1 - Ly_2 = 2t^3 - t^3 = t^3,  \tag 8$
so $2y_1 - y_2$ is a solution of (1); (4) is the correct choice.
