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Find the general solution of the differential equations:`(1+x^2)dy+2x y dx=cotx dx(x!=0)` |
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Answer» `(1+x^(2))dy+2xydx=cotxdx` `implies (dy)/(dx)+(2x)/(1+x^(2))y=(cotx)/(1+x^(2))` Here, `P=(2x)/(1+x^(2))` , `Q=(cotx)/(1+x^(2))` `:. I.F.=e^(int(2x)/(1+x^(2)))dx=e^(log(1+x^(2)))=(1+x^(2))` and general solution `y(1+x^(2))=int(cotx)/(1+x^(2))(1+x^(2))dx+c` `=intcotxdx+c` `implies y(1+x^(2))=logsinx+c` |
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