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The general solution of `e^(x) cos ydx-e^(x) sin dy=0` isA. `e^(x)cosy=k`B. `e^(x)siny=k`C. `e^(x)=k cos y`D. `e^(x)=k sin y` |
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Answer» Given that, `" "e^(x)cosydx-e^(x)sinydy=0` `rArr" "e^(x)cosydx=e^(x)sinydy` `rArr" "(dx)/(dy)=tany` `rArr" "dx=tanydy` On integrating both sides, we get `" "x=logsecy+C` `rArr" "x-C=logsecy` `rArr" "secy=e^(x-C)` `rArr" "secy=e^(x)e^(-C)` `rArr" "(1)/(cosy)=(e^(x))/(e^(C))` `rArr" "e^(x)cosy=e^(C)` `rArr" "e^(x)cosy=K" "` [where, `K=e^(C)`] |
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