Respuesta :
Answer:
y = [tex]\frac{3}{2 + 3Ce^{-3t} }[/tex]
Step-by-step explanation:
As given , y' = 3y - 2y²
⇒y' - 3y = -2y²
Divide by y² in the above equation
⇒[tex]\frac{y'}{y^{2} } - \frac{3y}{y^{2} } = -\frac{2y^{2} }{y^{2} }[/tex]
⇒[tex]\frac{y'}{y^{2} } - \frac{3}{y } = -2[/tex] ........(1)
Now , let [tex]\frac{1}{y}[/tex] = u
⇒-[tex]\frac{1}{y^{2} } \frac{dy}{dt} = \frac{du}{dt}[/tex]
⇒-[tex]\frac{1}{y^{2} }y' = \frac{du}{dt}[/tex]
∴ equation (1) becomes
-[tex]\frac{du}{dt}[/tex] - 3u = -2
⇒[tex]\frac{du}{dt}[/tex] + 3u = 2
It is a linear differential equation
Now,
Integrating factor = I.F = [tex]e^{\int\limits {3} \, dt }[/tex] = [tex]e^{3t}[/tex]
∴ The solution becomes
u.(I.F) = [tex]\int\limits {2.(I.F)} \, dt[/tex] + C
⇒u.([tex]e^{3t}[/tex]) = [tex]\int\limits {2.(e^{3t} )} \, dt[/tex] + C
⇒u.([tex]e^{3t}[/tex]) = [tex]\frac{2e^{3t} }{3}[/tex] + C
⇒u = [tex]\frac{2}{3} + Ce^{-3t}[/tex]
As [tex]\frac{1}{y}[/tex] = u
⇒[tex]\frac{1}{y}[/tex] = [tex]\frac{2}{3} + Ce^{-3t}[/tex]
⇒ y = [tex]\frac{1}{\frac{2}{3} + Ce^{-3t} } = \frac{3}{2 + 3Ce^{-3t} }[/tex]
⇒ y = [tex]\frac{3}{2 + 3Ce^{-3t} }[/tex]
