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.

Answer 1

\[y'=\frac{1-2t}{y}\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Yup. Is seperation of variables :3

Answer 6

now integrate both sides

Answer 7

but I keep getting the same answer ;c

Answer 8

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

Answer 9

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

Answer 10

\[y^2=2t-2t^2+C\]

Answer 11

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

Answer 12

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
\[y^2=2t-2t^2+4\]

Answer 13

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

Answer 14

Ah. plus minus. I forgot about that q-q

Thanks :3

Answer 15

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