Solve the initial-value problem for y\displaystyle {y}y as a function of x\displaystyle {x}x. (4−x2)dydx=1,y(0)=4\displaystyle {\left({4}-{x}^{{2}}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={1},{y}{\left({0}\right)}={4}(4−x2)dxdy=1,y(0)=4 y=\displaystyle {y}=y= Preview Question 6
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