Solve the initial value dydx 2xy2 0 y1 Solutiony 2xy2 0

Solve the initial value dy/dx + 2xy² = 0,

y(1) = ½

y\' +2xy^2 = 0

dy/y^2 = -2xdx

-1/y = -x^2 +c

x^2-1/y = c

y(1) = 1/2

1-2= c

c =-1

x^2 -1/y = -1

y =1/(x^2+1)

$Solve the initial value dy/dx + 2xy2 = 0, y(1) = ½Solutiony\' +2xy^2 = 0 => dy/y^2 = -2xdx => -1/y = -x^2 +c => x^2-1/y = c y(1) = 1/2 => 1-2= c =&g$

Solve the initial value dydx 2xy2 0 y1 Solutiony 2xy2 0

Solution

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