Find the general solution of the given differential

Given:
${\displaystyle 4y{}^{″}-4y^{\prime }-3=0}$
Concept:
The non ­- homogenous ODE, second-order linear equation of type ${\displaystyle ay{}^{″}+b{y}^{\prime }}$=constant
${\displaystyle y=y_h+y_p}$ is the general solution of ${\displaystyle ay{}^{″}+b{y}^{\prime }}$=constant
The solution:
${\displaystyle y_h}$ is the solution to the homogeneous ODE ${\displaystyle ay{}^{″}+b{y}^{\prime }=0}$
${\displaystyle y_p}$ the particular solution, is any function that satisfies the non-homogeneous equation
Solution:
${\displaystyle 4y{}^{″}-4y^{\prime }=0}$
The characteristic equation:
${\displaystyle 4m^2-4m=0}$
${\displaystyle 4m(m-1)=0}$
${\displaystyle m=0}$ and ${\displaystyle m=1}$
The solution when roots are ${\displaystyle a=0}$ and ${\displaystyle b=0}$ is ${\displaystyle c_1{e}^{ax}+c_2{e}^{bx}}$
${\displaystyle y_h=c_1{e}^{1x}+c_2{e}^{\left(0\right)x}}$
${\displaystyle y_h=c_1{e}^x+c_2{e}^0}$
${\displaystyle y_h=c_1{e}^x+c_2}$
Now,
${\displaystyle y_p=-\frac{3}{4}t}$
so,
${\displaystyle y=y_h+y_p}$
${\displaystyle y=\left(c_1{e}^x\right)+c_2)+(-\frac{3t}{4})}$
${\displaystyle y=c_1{e}^x+c_2-\frac{3t}{4}}$
Answer:
${\displaystyle y=c_1{e}^x+c_2-\frac{3t}{4}}$