Solving Problem 1: 3x' + 1/t x = t

[ollyfinn]

I have the following two questions to solve

Problem 1.

3x' + 1/t x = t

and

Problem 2.

x' + 1/t x = ln t

I have followed a method detailed in my textbook to try and get an answer for Problem 1 but am a bit unsure so if anyone can clarify my workings below before I spend time trying to solve Problem 2.

3x' + 1/t x = t

Fits the format dx/dt + g(t)x = f(t)

For the integrating factor I(t) = e^∫g(t) dt

∫1/t = ln t

e^ln t = t

Multiply both sides by I(t) so the equation becomes d/dt(I(t)x(t)) = I(t)f(t)

3 d/dt (tx) = t^2

Then I(t)x(t) = ∫I(t)f(t) dt + C

3tx = ∫t^2 dt + C

3tx = (t^3)/3 + C

x(t) = ((t^3)/3 + C)/3t

x(t) = 1/6t^2 + C1/3t^-1

Does this look right? If so I will attempt Problem 2. Thanks for any assistance.

 

[HallsofIvy]

I have the following two questions to solve

Problem 1.

3x' + 1/t x = t

and

Problem 2.

x' + 1/t x = ln t

I have followed a method detailed in my textbook to try and get an answer for Problem 1 but am a bit unsure so if anyone can clarify my workings below before I spend time trying to solve Problem 2.

3x' + 1/t x = t

Fits the format dx/dt + g(t)x = f(t)

For the integrating factor I(t) = e^∫g(t) dt

∫1/t = ln t

e^ln t = t

Multiply both sides by I(t) so the equation becomes d/dt(I(t)x(t)) = I(t)f(t)

3 d/dt (tx) = t^2

Then I(t)x(t) = ∫I(t)f(t) dt + C

3tx = ∫t^2 dt + C

3tx = (t^3)/3 + C

x(t) = ((t^3)/3 + C)/3t

x(t) = 1/6t^2 + C1/3t^-1

3*3 is NOT 6! Other than that, and the need for parentheses to make it clearer (most people would interpret "1/6t^2" as "1/(6t^2)" and I don't think that is what you mean), you are doing this correctly.

Does this look right? If so I will attempt Problem 2. Thanks for any assistance.

 

[ollyfinn]

You are right my parentheses does need to be clearer.

My final answer should be:

x(t) = 1/9(t^2) + C1/3(t^-1)

Does that look better?