+1313 Dm/dt=3m, m(1)=5
1/dt=3m/dm. Flip.
dt=1/3m*dm
● dt=● 1/3m*dm
t=ln|3m|
t+C=3ln|m|
e^t*e^C=3m
Ce^t=3m
1/3(Ce^t=3m)=m
1/3(C)*1/3e*....?
Going with my way please complete and show errors. Please also show what should have been done to iscolate the m before integrating and lastly if that is simpler than my solution. Answer is m=5e^3t-3. Thanks.
+4473 So, you have Dm/dt = 3m
Get m by itself -->
Dm = 3m dt --> Dm / m = 3 dt...go from there -->
ln |m| = 3t + C
m = e^(3t+C)
m = Ce^(3t)
5 = Ce^(3*1)
5 = Ce^(3)
C = 5 / e^3 --> m = 5e^(3t) / e^3 --> e^3 is same as e^-3 in the numerator -->
m = 5e^-3*e^3t --> Combine the e's -->
m = 5e^(3t-3).
+1313 
Am I correct now with my solution and method? Initial conditions are in the previous image I posted.
+1313 Why does the 4 not gwr integrated to have P attached and left as a constant?
+4473 Oh, because the integral of 1/x dx is ln(x) and it follows that the integral of 1/(x+2) dx is ln(x+2) -->
If we do u-substitution, u = x + 2 --> du = dx then 1/u du = ln u = ln(x + 2).
This situation is similar to the one in your question.
+1313 I am not sure what yo do here or if I was on the right path. Let me show you.
+4473 integral of 1/(0.3Q) dQ --> we can separate this into two fractions --> 1/(0.3)*(1/Q)
Now, the constant we can take out which becomes 3.33 and then 1/Q becomes ln(Q).
So, t + C + 120 = 3.33ln(Q) --> Divide both sides by 3.33.
t + C + 120 / (3.33) = ln(Q)
e^(t + C + 120 / (3.33)) = Q
50 = e^(0 + C + 120 / (3.33))
50 = e^(120 + C / (3.33))
ln(50) = ln (120 + C / (3.33))
ln(50) = ln (120 + C) - ln(3.33)
ln(50) + ln(3.33) = ln(120 + C)
Hmm, I am stuck here...I took exponents but didn't work out..
