We are given the differential equation:
‌=y+3along with the initial condition:
y(0)=2First, we need to solve the differential equation. Let's rewrite it as:
‌−y=3We recognize this as a first-order linear differential equation of the form:
‌+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−xWe multiply the original differential equation by the integrating factor:
e−x‌−e−xy=3e−xThis can be written as:
‌(ye−x)=3e−x We integrate both sides with respect to
x :
∫‌(ye−x)‌dx=∫3e−x‌dxSo, 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+CexUsing the initial condition
y(0)=2, we can find
C :
2=−3+Ce0 Since
e0=1 :
2=−3+CTherefore:
C=5 Substituting
C back into the solution for
y :
y=−3+5exNow, we need to find
y(log‌2) :
y(log‌2)=−3+5elog‌2Since
elog‌2=2 :
y(log‌2)=−3+5(2)=−3+10=7Therefore, the value of
y(log‌2) is:
Option D: 7