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Show that the general solution of the differentiaequation `(dy)/(dx)+(y^2y+1)/(x^2+x+1)=0`is given by `x+y+1=A(1-x-y-2x y)`where A is a parameter. |
Answer» Given differntial equation is : `(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0` `implies (dy)/(y^(2)+y+1)+(dx)/(x^(2)+x+1)=0` On integration `int(dy)/(y^(2)+y+1)+int(dx)/(x^(2)+x+1)=C` `impliesint(dy)/(y^(2)+y+1+((1)/(2))^(2)-((1)/(2))^(2))+int(dx)/(x^(2)+x+1+((1)/(2))^(2)-((1)/(2))^(2))=C` `impliesint(dy)/((y+(1)/(2))^(2)+(1-(1)/(4)))+int(dx)/((x+(1)/(2))^(2)+(1-(1)/(4)))=C` `impliesint(dy)/((y+(1)/(2))^(2)+((sqrt(3))/(2))^(2))+int(dx)/((x+(1)/(2))^(2)+((sqrt(3))/(2))^(2))=C` `implies(2)/(sqrt(3))tan^(-1)((y+(1)/(2))/((sqrt(3))/(2)))+(2)/(sqrt(3))tan^(-1)((x+(1)/(2))/((sqrt(3))/(2)))=C` `implies tan^(-1)(2y+1)/(sqrt(3))+tan^(-1)(2x+1)/(Sqrt(3))=(sqrt(3)C)/(2)=k` (say) `impliestan^(-1)[((2y+1)/(sqrt(3))+(2x+1)/(sqrt(3)))/(1-((2y+1)/(sqrt(3)))((2x+1)/(sqrt(3))))]=k` `impliestan^(-1)[((2y+1+2x+1)/(sqrt(3)))/(1-((4xy+2x+2y+1)/(3)))]=k` `implies(2sqrt(3)(x+y+1))/(3-(4xy+2x+2y+1))=tank` `implies(2sqrt(3)(x+y+1))/(2(1-x-y-2xy))=tank` `implies x+y+1=(1)/(sqrt(3))tank(1-x-y-2xy)` Let `A=(1)/(sqrt(3))tank` which is an arbitrary constant. `implies x+y+1=A(1-x-y-2xy)` |
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