Possible webpage title: Finding Solutions to ODEs with Separation of Variables

[Samantha24]

a) Derive a formula for solutions of the ode equation

b) If (t0, x0) lies in the subset of the (t,x)-plane given by x>1 write down a formula for the unique solution of the equation below through this point.

c)Give a formula for two solutions to equation below through the point (0,1) in the (t,x)-plane.

dx/dt = 6t*((x-1)^(2/3))a)

INT 1/(u^(2/3)) du = INT 6t dt, where u=(x-1)
3*(x-1^(1/3))+c = t^2 +c.
Hence x(t) = ((t^6)+27)/27

SOLVED; x(t)= (t^2+c)^2 +1

b) I subbed t=0 and x=0 into x(t) to work out the constant.

0=c^2 + 1
c= +i, -i

So, x(t)= (t^2-i)^2 +1 and x(t)= (t^2+i)^2 +1

Is this all I do? I don't really understand what it's asking me to do.

c) Confused.. I thought you sub t=0 and y=1 into x(t) but I get 1= +-i^2 +1 ??

 

[HallsofIvy]

Derive a formula for solutions of the ode equationdx/dt = 6t*((x-1)^(2/3))
INT 1/(u^(2/3)) du = INT 6t dt, where u=(x-1)
3*(x-1^(1/3))+c = t^2 +c.

= 3t^2+ c, not just t^2+ c. And, of course, you divide both sides by 3 to get
(x-1)^(1/3)= t^2+ c where I have combined the constants into (divided by 3) into one.

Now cube: x- 1= (t^2+ c)^3, x= (t^2+ c)^2+ 1.

Hence x(t) = ((t^6)+27)/27

When I differentiate x(t) I don't get the dx/dt above. Please help

 

[Samantha24]

AAH! Yeah I just realized my stupid error. Thank you for your help!

 

[vela]

a) Derive a formula for solutions of the ode equation

b) If (t0, x0) lies in the subset of the (t,x)-plane given by x>1 write down a formula for the unique solution of the equation below through this point.

c)Give a formula for two solutions to equation below through the point (0,1) in the (t,x)-plane.

dx/dt = 6t*((x-1)^(2/3))a)

INT 1/(u^(2/3)) du = INT 6t dt, where u=(x-1)
3*(x-1^(1/3))+c = t^2 +c.
Hence x(t) = ((t^6)+27)/27

SOLVED; x(t)= (t^2+c)^2 +1

Even with your edits, this solution is still incorrect. HallsofIvy made a typo, which you apparently didn't notice.

b) I subbed t=0 and x=0 into x(t) to work out the constant.

0=c^2 + 1
c= +i, -i

So, x(t)= (t^2-i)^2 +1 and x(t)= (t^2+i)^2 +1

Is this all I do? I don't really understand what it's asking me to do.

I'm not sure why you're using t=0 and x=0. The problem states that the solution passes through the point (t₀, x₀), which means x(t₀)=x₀. It also says this point lies in the half plane x>1, which means x can't be 0.

c) Confused.. I thought you sub t=0 and y=1 into x(t) but I get 1= +-i^2 +1 ??

There's no y in this problem. You mean x=1. If you fix part (a) and get the correct solution, this part should work out with the method you tried.

But now you need another solution. You might have noticed in part (b), the problem refers to a unique solution passing through a point whereas in part (c), it asks for two solutions that pass through (0,1). Note that the point (0,1) doesn't lie in the half plane x>1. Think about why this condition admits a second solution. Go back to the original equation, and in particular, look at the steps you used in part (a) to find x(t). Are those valid for the point (0,1)?