differential equation 42= 3(dT/dt) + 0.03T models temp T of

xc630

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The differential equaiton 42= 3(dT/dt) + 0.03T models the temperature T, above room temperature, of an electic burner. T is measured in degress Fahreneit and t is greather than or equal to zero is in seconds.

a) find the general solution of the diff equation in terms of a constant k.
b) when the burner is first turned on, T= 0 degress Fahrenheit. use this fact to find the value of k
c) The burner is unsafe if its temperature can exceed 1500 degrees fahrenheit. Is this burner safe or unsafe?

So far I have the following:

42 = 3(dT/dt) + 0.03T

42 dt = 3dT + 0.03 T

Taking the integral

42t = 3T + 0.03 + K

K= 42t/ (3T+0.03)

Then for part b:

k = 42t/ (3(0) +0.03)
What would little t be int his case?
I am also unsure what to do for part c.
Thanks for any help.
 
I believe you went about it a little wrong. Find your integrating factor.

\(\displaystyle \L\\3T'+\frac{3}{100}t=42\)

\(\displaystyle \L\\T'+\frac{1}{100}t=14\)

The integrating factor is \(\displaystyle \L\\e^{\int\frac{1}{100}dt}=e^{\frac{1}{100}t}\)

\(\displaystyle \L\\\frac{d}{dx}[Te^{\frac{1}{100}t}]=14e^{\frac{1}{100}t}\)

Inetgrate and get:

\(\displaystyle \L\\Te^{\frac{1}{100}t}=1400e^{\frac{1}{100}t}+K\)

\(\displaystyle \L\\\fbox{T=1400+Ke^{\frac{-1}{100}t}}\)

Now, use your initial condition to answer your questions. What is its maximum temperature?.
 
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