Differential Equation: y' = (1-2t)/y with initial Value of y(1) = -2

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

but it says my answer is wrong. I've redone it a few times, and can't pick up where I did wrong.

Question

Differential Equation: y' = (1-2t)/y with initial Value of y(1) = -2

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

but it says my answer is wrong. I've redone it a few times, and can't pick up where I did wrong.

myininaya · Answer

$$y'=\frac{1-2t}{y}$$

myininaya · Answer

looks like a separation of variables problem
i'm excited! :)

myininaya · Answer

$$\frac{dy}{dt}=\frac{1-2t}{y}$$

myininaya · Answer

multiply y on both sides
and multiply by dt on both sides to obtain
$$y dy =(1-2t) dt$$

anonymous · Answer

Yup. Is seperation of variables :3

myininaya · Answer

now integrate both sides

anonymous · Answer

but I keep getting the same answer ;c

myininaya · Answer

$$\int\limits_{}^{} y dy= \int\limits_{}^{}(1-2t) dt$$

myininaya · Answer

$$\frac{y^2}{2}=t-t^2+C$$

myininaya · Answer

$$y^2=2t-2t^2+C$$

myininaya · Answer

2 times a constant is still a constant so instead of 2c i just wrote c

myininaya · Answer

now if t=1 then y=-2
so apply initial condition we obtain
$$(-2)^2=2(1)-2(1)^2+C $$
4=2-2+C
4=C
so we have
$$y^2=2t-2t^2+4$$

myininaya · Answer

$$y=\pm \sqrt{2t-2t^2+4}$$

anonymous · Answer

Ah. plus minus.  I forgot about that q-q 

Thanks :3

myininaya · Answer

np
gj to you for doing everything correctly up into that point
lol
i mean you had one solution for y
thats real good