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solve the separable differential equation for u: du/dt=e^(6u+3t) use the following initial condition: u(0)=12 so u(t)=?
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if its separable; then separate :)
e^(a+b) = e^a e^b
is this a differential equations course?
e^6u+e^3t like that
not the "+" between it, that i can see
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so du/e^(6u)=e^(3t)+dt
there should be no "+" sign in either case lol
its just a chapter were going through for calculus 2
\[\int\limits_{}^{}\frac{1}{e^{6u}} du=\int\limits_{}^{}e^{3t}dt\]
so instead du/e^(6u)=e^(3t)(dt)
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\[\frac{du}{dt}=e^{6u+3t}\] \[\frac{du}{dt}=e^{6u}\ e^{3t}\] \[\int\{\frac{du}{e^{6u}}=e^{3t}dt\}\]
so ln(abs(e^6u))=e^3t^2+c
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