More Differential Probs: y'=y^2-4, y(0)=0; dy/dt=(1-2t)/y

[scrum]

Sorry I'm posting another thread so soon, I'm just stuck on these differential equations and my new proffessor has set the online homework system to only allow us to attempt to submit our answers three times, which is very frustrating.

I've got this one.

y' = y^2 - 4 , y(0) = 0

My work.

dy/dt = y^2 - 4

1 (dy) / (y^2 - 4) = (dt)

Integrate

(1/4)(ln(y-2)-ln(y+2)) = t + C

multiply both by 4

ln(y-2)-ln(y+2)) = 4t + C

(y-2) / (y+4) = Ce^4t

followed by a page of scrawl that is me and my friends attempts to solve that for y.

Also I think I have this one but I'm down to my last attempt so could you tell me if I am wrong

dy/dt = (1 - 2t)/y

y(dy) = (1-2t)(dt)

(y^2)/2 =t-t^2 + C

(y^2) = 2t - 2t^2 + C

y = sqrt(2t-2t^2 + C)

y(1) = -2

-2 = sqrt(c)

c = 4

y = - sqrt(2t-2t^2+4) ?

 

[Deleted member 4993]

Re: More Differential Problems

Sorry I'm posting another thread so soon, I'm just stuck on these differential equations and my new proffessor has set the online homework system to only allow us to attempt to submit our answers three times, which is very frustrating.

I've got this one.

y' = y^2 - 4 , y(0) = 0

My work.

dy/dt = y^2 - 4

1 (dy) / (y^2 - 4) = (dt)

Integrate

(1/4)(ln(y-2)-ln(y+2)) = t + C

multiply both by 4

ln(y-2)-ln(y+2)) = 4t + C

(y-2) / (y+4) = Ce^4t<<<< Correct

Now apply the initial condition (at t=0, y = 0) to find 'C' and you are done.
then to check your answer, - differentiate and plug it back into your original ODE - does it solve it?





followed by a page of scrawl that is me and my friends attempts to solve that for y.

Also I think I have this one but I'm down to my last attempt so could you tell me if I am wrong

dy/dt = (1 - 2t)/y

y(dy) = (1-2t)(dt)

(y^2)/2 =t-t^2 + C

(y^2) = 2t - 2t^2 + C

y = sqrt(2t-2t^2 + C)

y(1) = -2

-2 = sqrt(c)

c = 4

y = - sqrt(2t-2t^2+4)

To check - differentiate and plug it back into your original ODE - does it solve it?
?