InterviewSolution
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InterviewSolution
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Solve the differential equation `cos (x+y) dy=dx.` Hence find the particular solution for `x=0 and y=0.` |
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Answer» The given differential equartion is `(dy)/(dx)=(1)/(cos (x+y))" "...(1)` Let `x+y=v` `1+(dy)/(dx)=(dv)/(dx)` `(dy)/(dx)=(dv)/(dx)-1` Substituting `(dy)/(dx)` in equation (1), we get `(dv)/(dx)-1=(1)/(cos v)` `(dv)/(dx)=(1+cos v)/(cos v)` `therefore` Integrating `int(cos v dv)/(1+cos v)dv=intdx+c` `int(1+cos v)/(1+cos v)dv-int(1)/(a+cos v)dv=intdx+c` `int 1dv-int(1)/(2cos^(2)""(v)/(2))dv=x+c` `v-1/2intcos^(2)""(v)/(2)dv=x+c` `v-tan""(v)/(2)=x+c` `(x+y)-tan((x+y)/(2))=x+c" "...(2)` which is the required general solution of the given equation. For `x=0 and y=0,` equation (2), becomes, `(0+0)-tan((0)/(2))=0+c` `impliesc=0` Putting the value of c in equation (2), we get `x+y-tan((x+y)/(2))=x` `impliesy=tan ((x+y)/(2))` |
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