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The solution of differential equation `(1+y^(2))+(x-e^(tan^(-1)y))(dy)/(dx)=0`, isA. `xe^(2tan^(-1)y)=e^(tan^(-1)y+K)`B. `(X-2)=Ke^(tan^(-1)y)`C. `2xe^(tan^(-1)y)=e^(2 tan^(-1)y)+K`D. `xe^(tan^(-1)y)=tan^(-1)y+K` |
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Answer» Correct Answer - C We have, `(1+y^(2))+(x-e^(tan^(-1)))(dy)/(dx)=0` `rArr" "(dx)/(dxy)+(x)/(1+y^(2))=(e^(tan^(-1)y))/(1+y^(2))" ...(i)"` The linear differential equation with `"I.F."=e^(int(1)/(1+y^(2))dy)=e^(tan^(-1)y)` Multiplying (i) by I.F. and integrating, we get `xe^(tan^(-1)y)=int(e^(2tan^(-1)y))/(1+y^(2))dy` `rArr" "xe^(tan^(-1)y)=(1)/(2)inte^(2tan^(-1)y)d(2tan^(-1)y)` `rArr" "xe^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C` `rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1))+K," where K"=2C` |
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