Find constant solutions + Show every function

[PM123]

Hi, I'm having trouble with these two questions.


For a, I set dy/dt to 0, and solved for y and got 1/2. Since y = k, I think k=1/2?

For b, I'm confused about what this problem means. I derived to get dy/dt = C * (-2e^-2t)
I set dy/dt to 0, and get 0 = -2Ce^-2t, but then I solve for C to get C = 0, which seems weird as an answer? Not sure if I'm doing this right here.

I'd appreciate if someone could help/check my work. Thanks!

 

[MarkFL]

Please attach the image inline rather than posting a link to it. It's a great deal less convenient for our helpers if they have to follow a link, and many will simply refuse to do so. I tried editing your post to hotlink to the image, but apparently the image hosting site doesn't allow that.

 

[PM123]

Done.

 

[MarkFL]

For part b), we are told to let:

[MATH]y=\frac{1}{2}+Ce^{-2t}[/MATH]
And so we want to compute:

[MATH]\d{y}{t}=-2Ce^{-2t}[/MATH]
Now, substitute into the given ODE (for \(y\) and [MATH]\d{y}{t}[/MATH]) and confirm that an identity is the result.

 

[Steven G]

Hi, I'm having trouble with these two questions.
View attachment 12050

For a, I set dy/dt to 0, and solved for y and got 1/2. Since y = k, I think k=1/2?

For b, I'm confused about what this problem means. I derived to get dy/dt = C * (-2e^-2t)
I set dy/dt to 0, and get 0 = -2Ce^-2t, but then I solve for C to get C = 0, which seems weird as an answer? Not sure if I'm doing this right here.

I'd appreciate if someone could help/check my work. Thanks!

I think that you could have done this more easily by simply letting y=k (as instructed). So dy/dt = 0. Substituting yields 0= 1-2k. So yes, k=1/2.