Two Ways to Solve

[Jason76]

Solve the following initial value problem for \(\displaystyle u\) as a function of \(\displaystyle t\)

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle k\) and \(\displaystyle m\) are positive constants. \(\displaystyle u(0) = u_{0}\)

A. As a first order diff equation:

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

Integrating factor

\(\displaystyle e^{\int \dfrac{k}{m} (dt)} = e^{dfrac{k}{m}t }\)

\(\displaystyle \dfrac{du}{dt}(e^{\dfrac{k}{m}t}) + \dfrac{k}{m}u (e^{\dfrac{k}{m}t}) = 0 (e^{\dfrac{k}{m}t}) \)

\(\displaystyle \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t)] = 0\)

\(\displaystyle \int \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t) (dt)] = \int 0 (dt)\)

\(\displaystyle e^{\dfrac{k}{m}t})(t) = 0\) ?? How to get \(\displaystyle t\) by itself. Use natural log?

B. Using Separation of Variables

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle du + \dfrac{k}{m}u(dt) = 0\)

\(\displaystyle \dfrac{du}{u} + \dfrac{k}{m}(dt) = 0\)

\(\displaystyle \dfrac{du}{u} = - \dfrac{k}{m}(dt)\) ???

 

[Deleted member 4993]

Solve the following initial value problem for \(\displaystyle u\) as a function of \(\displaystyle t\)

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle k\) and \(\displaystyle m\) are positive constants. \(\displaystyle u(0) = u_{0}\)

A. As a first order diff equation:

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

Integrating factor

\(\displaystyle e^{\int \dfrac{k}{m} (dt)} = e^{dfrac{k}{m}t }\)

\(\displaystyle \dfrac{du}{dt}(e^{\dfrac{k}{m}t}) + \dfrac{k}{m}u (e^{\dfrac{k}{m}t}) = 0 (e^{\dfrac{k}{m}t}) \)

\(\displaystyle \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t)] = 0\)

\(\displaystyle \int \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t) (dt)] = \int 0 (dt)\)

\(\displaystyle e^{\dfrac{k}{m}t})(t) = \)0 ?? How to get \(\displaystyle t\) by itself. Use natural log?
B. Using Separation of Variables

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle du + \dfrac{k}{m}u(dt) = 0\)

\(\displaystyle \dfrac{du}{u} + \dfrac{k}{m}(dt) = 0\)

\(\displaystyle \dfrac{du}{u} = - \dfrac{k}{m}(dt)\) ???

Now integrate both sides...

ln(u) = -(k/m)*t +C₁

u = e^(C1) * e^[-(k/m)*t]

u = C* e^[-(k/m)*t]

.

 

[HallsofIvy]

Solve the following initial value problem for \(\displaystyle u\) as a function of \(\displaystyle t\)

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle k\) and \(\displaystyle m\) are positive constants. \(\displaystyle u(0) = u_{0}\)

A. As a first order diff equation:

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

Integrating factor

\(\displaystyle e^{\int \dfrac{k}{m} (dt)} = e^{dfrac{k}{m}t }\)

\(\displaystyle \dfrac{du}{dt}(e^{\dfrac{k}{m}t}) + \dfrac{k}{m}u (e^{\dfrac{k}{m}t}) = 0 (e^{\dfrac{k}{m}t}) \)

\(\displaystyle \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t)] = 0\)

\(\displaystyle \int \dfrac{d}{du}[e^{\dfrac{k}{m}t})(t) (dt)] = \int 0 (dt)\)

\(\displaystyle e^{\dfrac{k}{m}t})(t) = 0\) ?? How to get \(\displaystyle t\) by itself. Use natural log?

You mean \(\displaystyle e^{\frac{k}{m}t}= C\) where C can be any constant. To solve for t, yes, take the logarithm
of both sides: \(\displaystyle \frac{k}{m}t= C'= ln(C)\)

Using Separation of Variables

\(\displaystyle \dfrac{du}{dt} + \dfrac{k}{m}u = 0\)

\(\displaystyle du + \dfrac{k}{m}u(dt) = 0\)

\(\displaystyle \dfrac{du}{u} + \dfrac{k}{m}(dt) = 0\)

\(\displaystyle \dfrac{du}{u} = - \dfrac{k}{m}(dt)\) ???

Now integrate both sides. What is \(\displaystyle \int \frac{du}{u}\)? What is \(\displaystyle \int -\frac{k}{m} dt= -\frac{k}{m} \int dt\)?