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Solution of the differential equation `cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`isA. `tanx=(secx+c)y`B. `secx=(tanx+c)y`C. `ysec x=tanx+c`D. `ytanx =secx+c` |
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Answer» Correct Answer - B `cosxdy=y(sinx-y)dx` `rArr (dy)/(dx)=ytanx-y^(2)secx` `rArr 1/y^(2)(dy)/(dx)-1/ytanx=-secx` Let `1/y=t` `therefore -1/y^(2)(dy)/(dx)=(dt)/(dx)` `rArr -(dt)/(dx) -t tanx=-secx` `rArr-(dt)/(dx)+(tanx)t=secx` I.F. `=e^(int(tanxdx))=secx` The solution is t(I.F.) = `int(I.F.)secxdx` `rArr 1/ysecx=tanx+c` |
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