InterviewSolution
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InterviewSolution
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Find the general solution of the following differential equation :`(1y2)+((x-e^"tan"^((-1_"y")))dy)/(dx)=0` |
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Answer» Given, differential equation is `" "(1+y^(2))=(x-e^(tan^(-1)y)(dy)/(dx)` `rArr" "(1+y^(2))=-(x-e^(tan^(-1)y))(dy)/(dx)` `" "(1+y^(2))(dx)/(dy)=-x+e^(tan^(-1))y` `rArr" "(1+y^(2))(dx)/(dy)+x=e^(tan^(-1))y` `rArr" "(dx)/(dy)+(x)/(1+y^(2))=(e^(tan^(-1))y)/(1+y^(2))" "` [dividing throughout by `(1+y^(2))`] which is a linear differential equation. On comparing it with `(dx)/(dy)+Px=Q`, we get `" "P=(1)/(1+y^(2)),Q=(e^(tan^(-1)y))/(1+y^(2))` `" "IF=e^(intPdy)=e^(int(1)/(1+y^(2)))=e^(tan^(-1)y)` The general solution is `x*e^(tan^(-1)y)=int(e^(tan^(-1)y))/(1+y^(2))*e^(tan^(-1)y)dy+C` `rArr" "x*e^(tan^(-1)y)=int((e^(tan^(-1)y)))/(1+y^(2))*dy+C` Put `tan^(-1)y=trArr " "(1)/(1+y^(2))dy=dt` `therefore" "x*e^(tan^(-1)y)=inte^(2t)dt+C` `rArr" "x*e^(tan^(-1)y)=(1)/(2)e^(2tan^(-1)y)+C` `rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+2C` `rArr" "2xe^(tan^(-1)y)=e^(2tan^(-1)y)+K" "[becauseK=2C]` |
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