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Solution of the differential equation `cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`isA. `y tan x = sec x+C`B. `tan x= (sec x+C)y`C. `sec x =(tanx+C)y`D. `ysec x=tan x+C` |
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Answer» Correct Answer - C We have, `cosxdy=y(sinx-y)dx` `rArr" "(dy)/(dx)=y tan x-y^(2)secx` `rArr" "(1)/(y^(2))(dy)/(dx)+tanx(-(1)/(y))=-secx` `rArr" "(dv)/(dx)+(tanx)v=secx," where v"=-(1)/(y)` This is a linear differential equation with integrating factor `e^(inttanxdx)=secx.` So, its solution is given by `v sec x=-int sec^(2)xdx+C` `rArr" "v secx=-tanx-C` `rArr" "-(1)/(y)secx=-tanx-C` `rArr" "secx=y(tanx+C)` `ul("ALITER")` We have, `cosxdy=y(sinx-y)dx` `rArr" "cosxdy-y sinxdy=-y^(2)dx` `rArr" "(secxdy-y tanx sec x dx)/(y^(2))=-sec^(2)xdx` `rArr" "d((secx)/(y))=sec^(2)xdx` `rArr" "(secx)/(y)=tanx+C` `rArr" "secx=y(tanx+C)` |
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