=y+3 along with the initial condition: y(0)=2 First, we need to solve the differential equation. Let's rewrite it as: ‌
dy
dx
−y=3 We recognize this as a first-order linear differential equation of the form: ‌
dy
dx
+P(x)y=Q(x) where P(x)=−1 and Q(x)=3. The integrating factor µ(x) is given by: µ(x)=e∫P(x)‌dx=e−∫1‌dx=e−x We multiply the original differential equation by the integrating factor: e−x‌
dy
dx
−e−xy=3e−x This can be written as: ‌
d
dx
(ye−x)=3e−x We integrate both sides with respect to x : ∫‌
d
dx
(ye−x)‌dx=∫3e−x‌dx So, we get: ye−x=−3e−x+C where C is the constant of integration. Multiplying both sides by ex to solve for y : y=−3+Cex Using the initial condition y(0)=2, we can find C : 2=−3+Ce0 Since e0=1 : 2=−3+C Therefore: C=5 Substituting C back into the solution for y : y=−3+5ex Now, we need to find y(log‌2) : y(log‌2)=−3+5elog‌2 Since elog‌2=2 : y(log‌2)=−3+5(2)=−3+10=7 Therefore, the value of y(log‌2) is: Option D: 7