Solve dy/dx = y(4-y) given (0,1)

[burgerandcheese]

I just need help with 8(ii). Why is my working wrong? And why did my teacher say that there is no need for the modulus?
If there was no modulus, then I would arrive at -1/3 = e^(4c) after substituting (0,1) but then that would be a false since the range of the exp function is >0 ?? Please help me.

 

[Deleted member 4993]

View attachment 20176

I just need help with 8(ii). Why is my working wrong? And why did my teacher say that there is no need for the modulus?
If there was no modulus, then I would arrive at -1/3 = e^(4c) after substituting (0,1) but then that would be a false since the range of the exp function is >0 ?? Please help me.

View attachment 20177

Please show intermediate steps between steps 2 and 3.

 

[yoscar04]

You should get y(x)=4e^x/(e^(4x)+3).
You got a problem with one of your signs.

 

[LCKurtz]

A much less error prone simplification starts where you have [MATH]\frac y {y-4} = \pm e^{4x+4c}[/MATH]. Write that as [math]\frac y {y-4} = \pm e^{4x}e^{4c} = Ke^{4x}[/MATH] so [math]\frac y {y-4} = Ke^{4x}[/MATH]. Now put [MATH]y=1[/MATH] when [MATH]x=0[/MATH] and follow through. It won't be full of logs and will be easier to work without mistakes.

 

[yoscar04]

You should get y(x)=4e^x/(e^(4x)+3).
You got a problem with one of your signs.

forgot a 4 in the exponential in the numerator.

 

[Steven G]

When you multiply C by 4 you get C (it's a different C!).

C and 4C are just constants. Call them C.

 

[Steven G]

Note that y/(y-4) = (y-4)/(y-4) + 4/(y-4) = 1 + 4/(y-4).


Then solve for 1/(y-4) and easily solve for y.

Since e^a+b = e^ae^b you should realize that e^4x+c e^ce^4x=ce^4x

 

[Steven G]

Your 2nd error is in solving 1/3 = e^4c. You should get 4c = -ln(3)

 

[burgerandcheese]

You should get y(x)=4e^x/(e^(4x)+3).
You got a problem with one of your signs.

I followed Jomo's advice and got this. I think this is correct too.

A much less error prone simplification starts where you have [MATH]\frac y {y-4} = \pm e^{4x+4c}[/MATH]. Write that as [math]\frac y {y-4} = \pm e^{4x}e^{4c} = Ke^{4x}[/MATH] so [math]\frac y {y-4} = Ke^{4x}[/MATH]. Now put [MATH]y=1[/MATH] when [MATH]x=0[/MATH] and follow through. It won't be full of logs and will be easier to work without mistakes.

Thank you. Yes I'm aware of this method too, but I purposely chose to deal with logs for this question.

Please show intermediate steps between steps 2 and 3.

How I integrated 1/(y(4-y))?

 

[Steven G]

Now just follow Prof Khan's post and fix the 1st error and you'll be fine.

Why do you keep solving for c when you need 4c? Why not use c for 4c? Math has some grammar and it is grammatically correct to say c rather than 4c