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Solve the differential equation `"dy=cos x(2-y cosec x)dx"` given that `y=2, "when x" d=(pi)/(2)` |
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Answer» Given differential equation, `" "dy=cosx(2-y" cosec"x)dx` `rArr" "(dy)/(dx)=cosx(2-y" cosec"x)` `rArr" "(dy)/(dx)=2cosx-y" cosec"x*cosx` `rArr" "(dy)/(dx)=2cosx-ycotx` `rArr" "(dy)/(dx)+ycotx=2cosx` which is linear differential equation. On comparing it with `(dy)/(dx)+Py=Q`, we get `" "P=cotx, Q=2cosx` `" "IF=e^(intPdx)=e^(intcotxdx)=e^(logsinx)=sinx` The general solution is `" "y*sinx=int2cosx*sinxdx +C` `rArr" "y*sinx=intsin2xdx+C" "[because sin2x=2sinxcosx]` `rArr" "y*sinx=-(cos2x)/(2)+C" "...(i)` When `x=(pi)/(2) and y=2`, then `" "2*sin""(pi)/(2)=-(cos(2xx(pi)/(2)))/(2)+C` `rArr" "2*1=+(1)/(2)+C` `rArr" "2-(1)/(2)=CrArr(4-1)/(2)=C` `rArr therefore" "C=(3)/(2)` On subsituting the value of C in Eq. (i), we get `" "ysinx=-(1)/(2)cos2x+(3)/(2)` |
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