Question

Solve the initial-value problem y'=2y-y^2, y(0)=1.

Answer 1

Here's the work:

dy/dx=2y-y^2

dy/(2y-y^2)=dx

A/y+B/(2-y)=1

A=1/2, B=1/2,

(1/2)ln abs(y)+(1/2)ln abs(2-y)=x+c

sqrt(y)*sqrt(2-y)=x+c

sqrt(2y-y^2)=x+c

2y-y^2=(x+c)^2

-y^2+2y=(x+c)^2

-(y^2-2y)=(x+c)^2

(y^2-2y+1-1)=-(x+c)^2

(y-1)^2=1-(x+c)^2

y-1=+/-sqrt(1-(x+c)^2)

y=1+/-sqrt(1-(x+c)^2)

y(0)=1+/-sqrt(1-c^2)=1

Answer 2

The answer should be y=2/(1+e^(-2x)).

Answer 3

\[\int\frac{dy}{2-y}={-}\ln|2-y|+C\]

Answer 4

So you should have
\[\frac{1}{2}\ln|y|-\frac{1}{2}\ln|2-y|=\ln\sqrt{\frac{y}{2-y}}=x+C\]
which gives the general solution
\[\sqrt{\frac{y}{2-y}}=Ce^x~~\iff~~\frac{y}{2-y}=Ce^{2x}\]
Solving for \(y\):
\[\begin{align*}
y&=2Ce^{2x}-Cye^{2x}\\\\
y(1-Ce^{2x})&=2Ce^{2x}\\\\
y&=\frac{2Ce^{2x}}{1-Ce^{2x}}
\end{align*}\]

Answer 5

Let me check my work.

Answer 6

I see why. I got (1/2)ln abs(y)+(1/2)ln abs(2-y) but you got - instead of +, is it because of the - in 2-y?

Answer 7

Yes, you would have substituted \(u=2-y\), which would give \(-du=dy\).

Answer 8

So I got c=1/3.

Answer 9

Hold on, I think I made a mistake with the general solution...

Answer 10

I got y=2e^(2x)/(3-e^(2x)).

Answer 11

The
\[\begin{align*}\\
\frac{y}{2-y}&=Ce^{2x}\\\\
y&=Ce^{2x}-Cye^{2x}&\text{the 2 gets absorbed into the }C\\\\
y(1-Ce^{2x})&=Ce^{2x}\\\\
y&=\frac{Ce^{2x}}{1-Ce^{2x}}
\end{align*}\]
And so this gives
\[\begin{align*}
1&=\frac{C}{1-C}\\
1-C&=C\\
C&=\frac{1}{2}
\end{align*}\]
Hmm, WA says this answer is wrong. Let's start from the beginning...

Answer 12

\[\begin{align*}\frac{dy}{dx}&=2y-y^2\\
\frac{dy}{y(2-y)}&=dx\\
\int\frac{dy}{2y}+\int\frac{dy}{2(2-y)}&=\int dx\\
\frac{1}{2}\ln|y|-\frac{1}{2}\ln|2-y|&=x+C\\
\ln\left|\frac{y}{2-y}\right|&=2x+C\\
\frac{y}{2-y}&=Ce^{2x}\\
y&=2Ce^{2x}-yCe^{2x}\\
y+yCe^{2x}&=2Ce^{2x}\\
y(1+Ce^{2x})&=2Ce^{2x}\\
y&=\frac{2Ce^{2x}}{Ce^{2x}+1}
\end{align*}\]
I guess the 2 doesn't get absorbed into the \(C\)...

Now,
\[1=\frac{2C}{C+1}~~\iff~~C=1\]
So the solution is
\[\large y=\frac{2e^{2x}}{e^{2x}+1}\]

Answer 13

Finally getting the right answer! Wow, tough problem...Thanks a lot.

Answer 14

yw