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Find the particular solution of the differential equation `(1+e^(2x))dy+(1+y^2)e^x dx=0,`given that `y=1`when `x=0.` |
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Answer» `(1+e^(2x))dy+(1+y^(2))e^(x)dx=0` or `(dy)/(1+y^(3))+(e^(x)dx)/(1+e^(2x))=0` Integrating both sides, we get `int(dy)/(1+y^(2))+int(e^(x)dx)/(1+e^(2x))=C` `Now, tan^(-1)y+tan^(-1)(e^(x))=C`…………..(1) `therefore tan^(-1)+tan^(-1)=C` or `C=pi/2` Substituting `C=pi/2`, in equation(1), we get `tan^(-1)y+tan^(-1)(e^(x))=pi/2` This is the required solution of the given differential equation. |
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