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.
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.