Please help solve this question, Im having difficulty with it. Thanks.
$\displaystyle y_1'= 3y_1+ 4y_2$
$\displaystyle y_2'= -2y_1- y_2$
Differentiate the first equation to get $\displaystyle y_1"= 3y_1'+ 4y_2'$. From the second equation, $\displaystyle y_1"= 3y_1'+ 4(-2y_1- y_2)= 2y_1'- 8y_1- 4y_2$.
From the first equation again, $\displaystyle 4y_2= y_1'- 3y_1$ so $\displaystyle y_1"= 3y_1'- 8y_1- (y_1'- 3y_1)= 2y_1'- 5y_1$.
That is, $\displaystyle y_1"- 2y_1'+ 5y_1= 0$. Can you solve that?
That will have, of course, two constants in the general solution. To find $\displaystyle y_2$, use $\displaystyle 4y_2= y_1'- 3y_1$ so as not to introduce more constants.