Answer:
Solution: [tex]B=5+25e^{4-4t}[/tex]
Step-by-step explanation:
Given: [tex]\dfrac{dB}{dt}+4B=20[/tex]
with B(1)=30
The differential equation in form of linear differential equation,
[tex]\dfrac{dy}{dt}+Py=Q[/tex]
Integral factor, IF: [tex]e^{\int Pdt}[/tex]
General Solution:
[tex]y\cdot IF=\int Q\cdot IFdt[/tex]
[tex]\dfrac{dB}{dt}+4B=20[/tex]
P=4, Q=20
IF= [tex]e^{\int 4dt}=e^{4t}[/tex]
Solution:
[tex]Be^{4t}=\int 20e^{4t}dt[/tex]
[tex]Be^{4t}=5e^{4t}+C[/tex]
[tex]B=5+Ce^{-4t}[/tex]
B(1)=30 , Put t=1, B=30
[tex]30=5+Ce^{-4}[/tex]
[tex]C=25e^4[/tex]
[tex]B=5+25e^{4-4t}[/tex]
