dy/dx=2y^2 and if y=-1 when x=1, then when x=2, y=?
(you already posted this, but i still don't get it)
dy/dx = 2y^2
Integrating...y=2/3 y^3 + C
put 1,-1 into the equation, and solve for C
c=-1/3
so the equation is y=(2y^3)/3-1/3
then i have to find what y equals when x=2. how do i do that. there's no x in the equation.
Now, if the problem is dx/dy = 2y^2
then x=2/3 y^3 + C and you solve.
Recheck which it is.
To find the value of y when x = 2, we need to solve the given differential equation. Since we have the initial condition y = -1 when x = 1, we can use this condition to determine the value of the constant C.
The given differential equation is dy/dx = 2y^2. To solve this equation, we can separate the variables as follows:
1/(2y^2) dy = dx.
Integrating both sides, we get:
∫1/(2y^2) dy = ∫dx.
This can be simplified as:
-1/(2y) = x + K, where K is the constant of integration.
Rearranging the equation, we have:
y = -1/(2(x + K)).
Now, we can use the initial condition y = -1 when x = 1 to determine the value of the constant K:
-1 = -1/(2(1 + K)).
To solve for K, we can cross-multiply and simplify:
-2(1 + K) = 1,
-2 - 2K = 1,
-2K = 1 + 2,
-2K = 3,
K = -3/2.
Now we have determined the value of the constant K, so the equation becomes:
y = -1/(2(x - 3/2)).
To find the value of y when x = 2, we substitute x = 2 into the equation:
y = -1/(2(2 - 3/2)).
Simplifying further:
y = -1/(2(4/2 - 3/2)),
y = -1/(2(1/2)),
y = -1/(1),
y = -1.
Therefore, when x = 2, y is equal to -1.