Answer:
1. y = sin(x²+C)
2. see below
Step-by-step explanation:
1. ![\frac{dy}{dx} = 2x\sqrt{1-y^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%202x%5Csqrt%7B1-y%5E2%7D)
separation of variables
![\frac{dy}{\sqrt{1-y^2} } = 2xdx](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7B%5Csqrt%7B1-y%5E2%7D%20%7D%20%3D%202xdx)
integration both sides![\int\frac{dy}{\sqrt{1-y^2} } = \int2xdx](https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Bdy%7D%7B%5Csqrt%7B1-y%5E2%7D%20%7D%20%3D%20%5Cint2xdx)
you should get :
remember constant of integration!!
2. y = sin(x²+C)
3 = sin(0+C)
y(0) = 3 does not have a solution because our sin graph is not shifted vertically or multiplied by a factor whose absolute value is greater than 1, so our range is [-1,1] and 3 is not part of this range
Yes; this is a linear function because you plug in a number and multiply it by two, so the function would be y=2x, which is linear despite the fact that it lacks a b value.
Hope this helps! :)
Answer:
The answer to the question provided is 1.
Answer:
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This equation has two terms