Is my answer correct? Please check

[Math10]

Homework Statement

Solve Q'+(1/(50+t))Q=1.

Homework Equations

None.

The Attempt at a Solution

This is my work:
I know that the integrating factor is (1/(50+t)).
e^integral of (1/(50+t))dt=e^ln abs(50+t)=50+t
(50+t)Q'+Q=50+t
Q(50+t)=50t+t^2+C
Q=(50t+t^2+C)/(50+t)
I got Q=(50t+t^2+C)/(50+t) as the answer. Am I right?

 

[SteamKing]

You can check this yourself by substituting for Q and Q' in the original ODE.

 

[Ray Vickson]

Homework Statement

Solve Q'+(1/(50+t))Q=1.

Homework Equations

None.

The Attempt at a Solution

This is my work:
I know that the integrating factor is (1/(50+t)).
e^integral of (1/(50+t))dt=e^ln abs(50+t)=50+t
(50+t)Q'+Q=50+t
Q(50+t)=50t+t^2+C
Q=(50t+t^2+C)/(50+t)
I got Q=(50t+t^2+C)/(50+t) as the answer. Am I right?

I concur with SteamKing's answer. You can easily check it yourself, and it would be good to develop the habit of checking on your own.

 

[LCKurtz]

This appears to be a continuation of this thread:
https://www.physicsforums.com/threads/find-the-quantity-of-salt-in-the-tank.792459/

I think most of us at PF frown on abandoning a thread and then starting a new one like it was a different subject.

 

[Mark44]

I think most of us at PF frown on abandoning a thread and then starting a new one like it was a different subject.

Yes.

Math10, do not abandon a thread, and then start a new one on the same problem.

Also, as SteamKing and Ray Vickson have pointed out, you can and should, check answers to problems like this on your own.

 

[Math10]

Okay. I won't abandon a thread again.

 

[Math10]

And I have already solved the problem.