Question

Initial value problem, dy/dt + y/t = g(t), y(1)=1
No idea where to start.

Answer 1

Don't you find the general solution and then find the particular solution?

Answer 2

Actually I noticed something interesting...

Answer 3

I don't know how to begin getting the general solution, that's the problem. :(
All the questions we've done so far have had only one function in them.

For instance, I just solved \[dy/dt=e^{t-y}\] but that only has y as a function of t. This new question has both g AND y as functions of t.

Answer 4

\[

\begin{array}{rcl}
\frac{dy}{dt} + \frac{y}{t} &=& g(t) \\
\frac{dy}{dt}\cdot t + y\cdot 1 &=& t\cdot g(t) \\
\frac{d}{dt}(y\cdot 1)&=& t\cdot g(t)
\end{array}
\]

Answer 5

oops should be \[
\frac{d}{dt}\left( y\cdot t\right) =g(t)\cdot t
\]

Answer 6

By the way, do you know how to do something like the Laplace transform?

Answer 7

Anyway, when getting the general solution, It should be separable.

Answer 8

\[
\begin{array}{rcl}
\frac{dy}{dt} + \frac{y}{t} &=& 0 \\
\frac{dy}{dt} &=& - \frac{y}{t} \\
\frac{1}{y}\frac{dy}{dt} &=& - \frac{1}{t} \\
\int \frac{1}{y}\frac{dy}{dt} dt &=& \int - \frac{1}{t} dt \\
\int \frac{1}{y} dy &=& \int - \frac{1}{t} dt \\
\end{array}
\]

Answer 9

Never heard of the Laplace transformation, but that's sort of why I'm studying right now, haha

Answer 10

Anyway, see how the general solution is just \(\frac{C}{t}\)