Thread: first order differential equation

Solve for dy/dx+y/x=1/x

I'd like to see the correct way to solve it. I did it a ridiculous way and ended up with the answer, but I don't want to spend so much time next time.

General formula is y=e^-(P(x)dx)*int[(Q(x)e^(P(x)dx)dx], no?

P(x)y=y/x so P(x)=1/x
Q(x)=1/x so Q(x)=1/x as well
the integrating factor is = e^int[(P(x)dx)] = e^int[1/x] = e^[ln(x)] = x ... right?

I should just be able to plug in and solve, right?

Originally Posted by thedoge

Solve for dy/dx+y/x=1/x

I'd like to see the correct way to solve it. I did it a ridiculous way and ended up with the answer, but I don't want to spend so much time next time.

General formula is y=e^-(P(x)dx)*int[(Q(x)e^(P(x)dx)dx], no?

P(x)y=y/x so P(x)=1/x
Q(x)=1/x so Q(x)=1/x as well
the integrating factor is = e^int[(P(x)dx)] = e^int[1/x] = e^[ln(x)] = x ... right?

I should just be able to plug in and solve, right?

It's of variables seperable type.



so:





so:



RonL

RonL

Originally Posted by thedoge

Solve for dy/dx+y/x=1/x

This is first order linear differencial equation, solves easily.

You have,

We want to solve this on some open interval excluding zero.

The integrating factor is,

I will assume the interval is in that case,
.

Thus, the general solution is,
= .
Thus,
are all solutions.

What would need to be done additionally if you had initial conditions where x and y were constants?

Originally Posted by thedoge

What would need to be done additionally if you had initial conditions where x and y were constants?

The initial condition determine the arbitary constant, so if you have initial
condition at , you have:



or if you are ImPerfectHacker:



RonL

Originally Posted by thedoge

What would need to be done additionally if you had initial conditions where x and y were constants?

If the initial condition was,

Then, since,

We substitute,
and .
Thus,

Solve,

And have the solution is,

Which satisfies this differencial equation.