The homogeneous part of the ODE has characteristic equation
[tex]4y''+9y=0\implies 4r^2+9=0[/tex]
which has roots at [tex]r=\pm i\dfrac32[/tex]. This means the characteristic solution takes the form
[tex]y_c=C_1\cos\dfrac32x+C_2\sin\dfrac32x[/tex]
For the particular solution, we can attempt to find a solution of the form
[tex]y_p=a_0[/tex]
[tex]\implies {y_p}''=0[/tex]
and substituting into the nonhomogeneous ODE, we get
[tex]4(0)+9a_0=15\implies a_0=\dfrac{15}9=\dfrac53[/tex]
so that the particular solution is
[tex]y_p=\dfrac53[/tex]
and the general solution to the ODE is
[tex]y=y_c+y_p[/tex]
[tex]y=C_1\cos\dfrac32x+C_2\sin\dfrac32x+\dfrac53[/tex]
