Given the general solution to the differential equation, find the particular solution to the initial value problem.

a$.y=4x^2+C$ for $y^{\text{'}}=6x,y(2)=0$

b$.y=\frac{1}{1+C{e}^{-x}}$ for $y^{\text{'}}=y-y^2,y(0)=-\frac{1}{4}$

c$.x=C_{1}cost+C_{2}sint$ for $x^{\text{'}\text{'}}+x=0,x(0)=-1,x^{\text{'}}(0)=3$

d$.x=C_{1}cost+C_{2}sint$ for $x^{\text{'}\text{'}}+x=0,x(\frac{}{3})=\frac{\sqrt[\phantom{2}]{3}}{2},x^{\text{'}}(\frac{}{3})=0$

e$.y=C_1{e}^x+C_2{e}^{-x}$ for $y^{\text{'}\text{'}}-y=0,y(0)=1,y^{\text{'}}(0)=3$