There is a series of ODE problems I can't understand

[kikko]

Homework Statement

The problems are these:

y' + (3y/t) = (Sin(t)/t^3)
ty'-2y = t^3 + t^2, t>0
(general case)

y't^3+(3yt^2), y(2) = 0
(specific case)

Homework Equations

Basic ODE solving skills

The Attempt at a Solution

I can't figure out how to make the y's and y''s go on one side, and make the t's go on the other side of the equation. I think there is some completing a square trick or something to solve those.


Now, I did this problem earlier:
y + 3y = te^-3t

and integrated and multiplied both sides by two to get an implicit solution of:

2y+6y^2 = (-2/9)e^-3x (3x+1) + C

Did I do this correctly? As far as I now, all I have to do is integrate both sides (even though I don't see any dy's or dx's, so am confused).

 

[cragar]

y' + (3y/t) = (Sin(t)/t^3). on this problem try multiplying by an integrating factor.
look carefully at the left hand side, what would I have to multiply by to run the product rule backwards.

 

[kikko]

I remember integrating factors, but I don't see one I can apply.

Can we walk through it step by step?

the y' + (3y/t) = (sin(t)/(t^3))

I can't see how to solve this through Exact Equations methods or Integrating Factors methods.

 

[LCKurtz]

Now, I did this problem earlier:
y + 3y = te^-3t

and integrated and multiplied both sides by two to get an implicit solution of:

2y+6y^2 = (-2/9)e^-3x (3x+1) + C

Did I do this correctly?

I presume you mean the equation was $y'+3y=te^{-3t}$. But no! That is most emphatically not correct. Let me count the ways:
1. The answer would be a function of t, not x.
2. y is an unknown function of t. You can't find its antiderivative that way. If it happened that $y = \sin(t)$, would you claim its antiderivative was $\frac{\sin^2t} 2$?
3. Even if you could the antiderivative of 3y would be $\frac{3y^2} 2$ and when you multiplied it by 2 you wouldn't get $6y^2$.
I'm thinking you best read your text about the use of integrating factors for first order linear equations.

 

[kikko]

I see how that was wrong. I can't find in my textbook how to do problems like the ones I have listed. It's the Boyce ODE textbook. I'd likely be able to solve them all if we walked through 1 step by step.

 

[LCKurtz]

I see how that was wrong. I can't find in my textbook how to do problems like the ones I have listed. It's the Boyce ODE textbook. I'd likely be able to solve them all if we walked through 1 step by step.

I have the Boyce & DiPrima "Elementary Differential Equations" 6th edition sitting in front of me. In section 2.1 on page 19 of this edition it explains the integrating factor theory. Then examples 2 and 3 on pages 20-22 give worked examples. Have you looked at those?

 

[HallsofIvy]

Homework Statement

The problems are these:

y' + (3y/t) = (Sin(t)/t^3)

Multiply both sides by t to get ty'+ 3y= sin(t)/t^2.
That's an "Euler type" or "equipotential" equation and the change of variable x= ln(t) will convert it to an equation with constant coefficients.

ty'-2y = t^3 + t^2, t>0

this is, as it stands, an equipotential equation.

(general case)

y't^3+(3yt^2), y(2) = 0
(specific case)

I assume you mean t^3y'+ 2t^2y= 0. Dividing both sides by t^2 gives ty'+ 2y= 0, again an equipotential equation. In all of these, the substitution x= ln(t) gives an equation with constant coefficients.

Homework Equations

Basic ODE solving skills

The Attempt at a Solution

I can't figure out how to make the y's and y''s go on one side, and make the t's go on the other side of the equation. I think there is some completing a square trick or something to solve those.


Now, I did this problem earlier:
y + 3y = te^-3t

and integrated and multiplied both sides by two to get an implicit solution of:

2y+6y^2 = (-2/9)e^-3x (3x+1) + C

Did I do this correctly? As far as I now, all I have to do is integrate both sides (even though I don't see any dy's or dx's, so am confused).

 

[LCKurtz]

Multiply both sides by t to get ty'+ 3y= sin(t)/t^2.
That's an "Euler type" or "equipotential" equation and the change of variable x= ln(t) will convert it to an equation with constant coefficients.

True enough, but given the OP's apparent lack of sophistication with solution methods, woudn't you think he should first master the first order integrating factor method? He's likely to have more trouble with the $x=\ln t$ change of variables than with the solution itself.