Separable Differential Equations: dy/dx = x^4(y-2)

[aa1234aa]

Consider the differential equation dy/dx = x^4(y-2) and find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 0.

I got ln|y-2| = x^5/5 + c and did y-2 = ce^(x^5/5) before putting in the given values to get c = -2. Then I got the solution as y=2-2e^(x^5/5).

However, my teacher wants me to find c right after getting ln|y-2| = x^5/5 + c. This seems to lead to a different c-value and solution.

ln|0-2| = c
ln2 = c
ln|y-2| = x^5/5 + ln2
e ln|y-2| = e^(x^5/5) * e^ln2
y=2+2e^(x^5/5)

The online calculator says that 2-2e^(x^5/5) is the correct answer. How can I find c as my teacher wants me to, and what's wrong with the above?

 

[HallsofIvy]

Consider the differential equation dy/dx = x^4(y-2) and find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 0.

I got ln|y-2| = x^5/5 + c and did y-2 = ce^(x^5/5) before putting in the given values to get c = -2. Then I got the solution as y=2-2e^(x^5/5).

However, my teacher wants me to find c right after getting ln|y-2| = x^5/5 + c. This seems to lead to a different c-value and solution.

ln|0-2| = c
ln2 = c
ln|y-2| = x^5/5 + ln2
e ln|y-2| = e^(x^5/5) * e^ln2

Your error is in this step. Taking the exponential on both sides gives
|y- 2|= 2e^(x^5/5). That can then give either
y- 2= 2e^(x^5/5) or -(y- 2)= 2e^(x^5/5)
The condition that y(0)= 0 requires -(y- 2)= 2e^(x^5/5) not y- 2= 2e^(x^5/5).

y=2+2e^(x^5/5)

The online calculator says that 2-2e^(x^5/5) is the correct answer. How can I find c as my teacher wants me to, and what's wrong with the above?

 

[Ishuda]

Consider the differential equation dy/dx = x^4(y-2) and find the particular solution y = f(x) to the given differential equation with the initial condition f(0) = 0.

I got ln|y-2| = x^5/5 + c and did y-2 = ce^(x^5/5) before putting in the given values to get c = -2. Then I got the solution as y=2-2e^(x^5/5).

However, my teacher wants me to find c right after getting ln|y-2| = x^5/5 + c. This seems to lead to a different c-value and solution.

ln|0-2| = c
ln2 = c
ln|y-2| = x^5/5 + ln2
e ln|y-2| = e^(x^5/5) * e^ln2
y=2+2e^(x^5/5)

The online calculator says that 2-2e^(x^5/5) is the correct answer. How can I find c as my teacher wants me to, and what's wrong with the above?

BTW: As Halls of Ivy points out, your solution is correct, i.e.
-(y- 2)= 2e^(x^5/5) \(\displaystyle \implies\) y = 2 - 2e^(x^5/5).
It is, IMO, just that your teacher wants you to justify that the proper choice of |y-2| is -(y-2).

 

[aa1234aa]

Thank you both!

One quick question. Do I always have to plug in the initial condition to figure out whether to use (y - 2) or -(y - 2)?