InterviewSolution
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InterviewSolution
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Find the general solution of the differential equation `(1+tany)(dx-dy)+2x dy=0`A. `x(sin y + cos y)=sin y + ce^(y)`B. `x(sin y + cos y) = sin y + ce^(-y)`C. `y(sin x + cos x)=sin x + ce^(x)`D. None of these |
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Answer» Correct Answer - B We have, `(1 + tan y)(dx - dy) + 2x dy = 0` `rArr" "(1+tan y)dx =(1+tan y - 2x)dy` `rArr" "(dx)/(dy)+(2)/(1+tan y)x=1`, which is linear differential equation. I.F.`=e^(2int(dy)/(1+tan y))=e^(int(2 cos y)/(sin y + cos y)dy)` `=e^(int(1+(cos y-sin y)/(sin y + cos y))dy)=e^(y+log(cos y + sin y))` `=(cos y + sin y)e^(y)` So, the solution is `Xe^(y)(sin y + cos y) = int e^(y) (sin y + cos y) dy + c` `i.e. xe^(y)(sin y + cos y)=e^(y) sin y + c`. `i.e. x(sin y + cos y)=sin y + ce^(-y)` |
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