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Solve the initial value problem:
dy——— = 2xy², y = 2, when x = – 1. dxSeparate the variables in the equation above:

Integrate both sides:


Take the reciprocal of both sides, and then you have

In order to find the value of
C₁ , just plug in the equation above those known values for
x and
y, then solve it for
C₁:
y = 2, when
x = – 1. So,


Substitute that for
C₁ into (i), and you have

So
y(– 2) is

I hope this helps. =)
Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>
Answer:
nope
Step-by-step explanation:
it acutally equals to 292 I am sure!
Answer:
After 1 second, the ball will reach a maximum height of 16 feet
Step-by-step explanation:
The height of the ball after t seconds: h(t) = -16t^2 + 32t
The graph of this quadratic function is parabola which opens downwards. The vertex of a quadratic equation is the maximum or minimum point on the equation's parabola
t = -b/2a = -(32)/2(-16) = -32/-32 = 1 second
then
h(t) = -16(1)^2 + 32(1) = -16 + 32 = 16
After 1 second, the ball will reach a maximum height of 16 feet
Answer:
The corrrect answer is -16 and +3