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If `(2+sinx)(dy)/(dx)+(y+1)cosx=0` and `y(0)=1`, then `y((pi)/(2))` is equal toA. `+(1)/(3)`B. `(4)/(3)`C. `(1)/(2)`D. `-(2)/(3)` |
Answer» Correct Answer - C The given differential equation is `(2+sinx)(dy)/(dx)+(y+1)cosx=0` `rArr" "(2+sinx)dy(y+1)cosx dx =0` `rArr" "(1)/(y+1)dy+(cosx)/(2+sinx)dx=0` Integrating both sides, we get `log(y+1)+log(2+sinx)=logC` `rArr" "(y+1)(2+sinx)=C" ...(i)"` It is given that `y(0)=1` i.e. y = 1 when x = 0. Putting x = 0, y = 1 in (i), we get `2(2+0)=CrArr C=4` Putting C = 4 in (i), we obtain `(y+1)(2+sinx)=4" ...(ii)"` Putting `x=(pi)/(2)` in (ii), we get `y=(1)/(3)`. |
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