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Solve this differential equation dT/dt = -3/25(T-25)
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\[\frac{dT}{T-25}=-\frac{3}{25}dt\]integrate both sides\[\ln(T-25)=-\frac{3}{25}t\]\[\large{e^{\ln(T-25)}=e^{-\frac{3}{25}t}}\]
right lana :)
\[\large{T-25=e^{-\frac{3}{25}t}}\]
ah thank you ! :)
theres a +C after e^{3/2t}
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forgot it -.- one min let me type the answer
\[\large{T=Ce^{-\frac{3}{25}t}+25}\]
just integrate both the sides
because when integrating\[\ln(T-25)=-\frac{3}{25}t+C\]so when u take e of both sides\[\large{e^{\ln(T-25)}=e^{-\frac{3}{25}t+C}}\]\[\large{\(T-25)=e^{-\frac{3}{2}t}e^c}\]e^c is just a constant so u can write it as C or any letter
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