Question Solved1 Answer dy dt = Method of separation of variables for = f(t)g(y). Solve the following differential equations, with initial conditions when given. Unless noted, y = y(x). (b) xy' = (1 - y2)1/2 dy = xy), y(0) = 1 dc dy (d) = xy3, y(0) = 0 dx

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Transcribed Image Text: dy dt = Method of separation of variables for = f(t)g(y). Solve the following differential equations, with initial conditions when given. Unless noted, y = y(x). (b) xy' = (1 - y2)1/2 dy = xy), y(0) = 1 dc dy (d) = xy3, y(0) = 0 dx
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Transcribed Image Text: dy dt = Method of separation of variables for = f(t)g(y). Solve the following differential equations, with initial conditions when given. Unless noted, y = y(x). (b) xy' = (1 - y2)1/2 dy = xy), y(0) = 1 dc dy (d) = xy3, y(0) = 0 dx
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(b){:[xy^(')=(1-y^(2))^(1//2)],[(xdy)/(dx)=sqrt(1-y^(2))],[(dy)/(sqrt(1-y^(2)))=(dx)/(x)quad[1(dx)/(sqrt(1-x^(2)))=sin^(-1)x]]:}integrate 6oth side{:[int(dy)/(sqrt(1-y^(2)))=int(dx)/(x)quad[[c=" integrat "],[" const. "]]],[sin^(-1)y=ln |x|+c],[y=sin(ln(x)+c)]:}(c) The give sifferential equation is{:[(dy)/(dx)=xy^(3);y(0)=1],[int(dy)/(y^(3))=int xdx],[int y-3dy=int xdx],[(y-3+1)/(-3+1)=(x^(2))/(2)+c],[(1)/(-2y^(2))=(x^(2))/(2)+c]:}It is given that y(0)=1 i.e y=1So ea(0 (1) become{:[-(1)/ ... See the full answer