2(x+2)2f(x)−3(x+2)2=10∫0x(t+2)f(t)dt Differentiating both side $\begin{aligned}
& 4(x+2) f(x)+2 f(x)(x+2)^2-6(x+2)=10(x+2) f(x) \
& =(x+2) \frac{d y}{d x}-3 y=3 \
& \frac{1}{3} \int \frac{d y}{y+1}=\int \frac{d x}{x+2} \
& \ln |y+1=3 \ln | x+2 \mid+\ln c \
& y+1=(x+2)^3 c \
& \because y(0)=\frac{3}{2} \
& \Rightarrow \frac{5}{16}=c \
& \therefore \quad y=\frac{5}{16}(x+2)^3-1 \
& y(2)=\frac{5}{16} \times 64-1=19
\end{aligned}$