# Homework Help: First-Order Equations problem

1. Jan 14, 2010

### strawburry

1. The problem statement, all variables and given/known data

Consider the equation y' = f(at + by + c)
where a, b, and c are constants. Show that the substitution x = at + by + c
changes the equation to the separable equation x' = a + bf(x).
Use this method to find the general solution of the equation y' = (y+t)^2

2. Relevant equations

n/a

3. The attempt at a solution

not sure where to begin :/

2. Jan 14, 2010

### rock.freak667

so x = at + by + c

x'=?

3. Jan 14, 2010

### strawburry

mm so plug in the first equation into where by is?

4. Jan 14, 2010

### Dick

Differentiate x with respect to t and solve for y' in terms of x'. Then put that expression for y' into your original equation. Now everything is in terms of x.

5. Jan 14, 2010

### strawburry

thanks! i got how to do the first part, showing the substitution to show separable equation..

How do i continue on to second part of the question?? SOrryy im kinda slow :(

6. Jan 14, 2010

### Dick

7. Jan 14, 2010

### strawburry

x' = a + by' ??

8. Jan 14, 2010

### Dick

Sure. Now solve that for y' and put it back into your original equation to eliminate y.

9. Jan 14, 2010

### strawburry

y' = ( x' - a ) / b

then f(x) = (x'-a)/b

then x' = a + bf(x)

10. Jan 14, 2010

### Dick

Well, that's that then, right?

11. Jan 14, 2010

### strawburry

still not understanding
Use this method to find the general solution of the equation y' = (y+t)^2

12. Jan 14, 2010

### Dick

Try it. Substitute x=y+t. Do the same thing you just did to get an equation for x. What is it?

13. Jan 14, 2010

### strawburry

x = y + t
x' = y' +1

y= x^2...

x'-1= x^2

14. Jan 14, 2010

### Dick

Nice. Ok, so dx/dt=(1+x^2). That's a separable ode.

15. Jan 14, 2010

### strawburry

integral ( -x^2 dx) = integral (1)

- 1/3 (x^3) = t

x = cubed root (-3t) ?

16. Jan 14, 2010

### Dick

Oh, come on, that's just silly. dx/(1+x^2)=dt. Integrate both sides. I was sort of hoping you knew separable ODE's. Heard of them?

17. Jan 14, 2010

### strawburry

haha just started learning them :P

18. Jan 14, 2010

### Dick

Now's the time to use them.

19. Jan 14, 2010

tytyty