Question

Find the general solution of the following equation: y'(t) = 3y -5

Answer 1

Answer:

The general solution of the equation is y = $\frac{A}{3}e^{3t}$ + 5

Step-by-step explanation:

Since the differential equation is given as y'(t) = 3y -5

The differential equation is re-written as

dy/dt = 3y - 5

separating the variables, we have

dy/(3y - 5) = dt

dy/(3y - 5) = dt

integrating both sides, we have

∫dy/(3y - 5) = ∫dt

∫3dy/[3(3y - 5)] = ∫dt

(1/3)∫3dy/(3y - 5) = ∫dt

(1/3)㏑(3y - 5) = t + C

㏑(3y - 5) = 3t + 3C

taking exponents of both sides, we have

exp[㏑(3y - 5)] = exp(3t + 3C)

3y - 5 = $e^{3t}e^{3C}$        

3y - 5 = $Ae^{3t}$                $A = e^{3C}$

3y = $Ae^{3t}$ + 5    

dividing through by 3, we have

y = $\frac{A}{3}e^{3t}$ + 5

So, the general solution of the equation is y = $\frac{A}{3}e^{3t}$ + 5