Respuesta :
[tex]\dfrac{\mathrm dA}{\mathrm dt}=\dfrac{3\text{ gal}}{1\text{ min}}\dfrac{4\text{ lb}}{1\text{ gal}}-\dfrac{2\text{ gal}}{1\text{ min}}\dfrac{A\text{ lb}}{700+(3-2)t\text{ gal}}[/tex]
[tex]\iff A'+\dfrac2{700+t}A=12[/tex]
[tex](700+t)^2A'+2(700+t)A=12(700+t)^2[/tex]
[tex]((700+t)^2A)'=12(700+t)^2[/tex]
[tex](700+t)^2A=4(700+t)^3+C[/tex]
[tex]A=4(700+t)+\dfrac C{(700+t)^2}[/tex]
Initially, the tank contains 50 lbs of salt in 700 gal of solution, i.e. [tex]A(0)=\dfrac{50}{700}=\dfrac5{70}[/tex], so
[tex]\dfrac5{70}=2800+\dfrac C{700^2}[/tex]
Use this to solve for [tex]C[/tex].
[tex]\iff A'+\dfrac2{700+t}A=12[/tex]
[tex](700+t)^2A'+2(700+t)A=12(700+t)^2[/tex]
[tex]((700+t)^2A)'=12(700+t)^2[/tex]
[tex](700+t)^2A=4(700+t)^3+C[/tex]
[tex]A=4(700+t)+\dfrac C{(700+t)^2}[/tex]
Initially, the tank contains 50 lbs of salt in 700 gal of solution, i.e. [tex]A(0)=\dfrac{50}{700}=\dfrac5{70}[/tex], so
[tex]\dfrac5{70}=2800+\dfrac C{700^2}[/tex]
Use this to solve for [tex]C[/tex].