InterviewSolution
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InterviewSolution
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Solve the following differential equation: `(x^2dy)/(dx)=x^2+x y+y^2` |
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Answer» Given that, `" "x^(2)(dy)/(dx)=x^(2)+xy+y^(2)` `rArr" "(dy)/(dx)=1+(y)/(x)+(y^(2))/(x^(2))" "...(i)` Let `" "f(x,y)=1+(y)/(x)+(y^(2))/(x^(2))` `" "f(lamdax,lamday)=1+(lamday)/(lamdax)+(lamda^(2)y^(2))/(lamda^(2)x^(2))` `" "f(lamdax, lamday)=lamda^(0)(1+(y)/(x)+(y^(2))/(x^(2)))` `" "=lamda^(0)f(x,y)` which is homogeneous expression of degree 0. Put `" "y=vxrArr=(dy)/(dx)=v+x(dv)/(dx)` On substituting these values in Eq. (i), we get `" "(v+x(dv)/(dx))=1+V+V^(2)` `rArr" "x(dv)/(dx)=1+v+v^(2)-v` `rArr" "x(dv)/(dx)=1+v^(2)` `rArr" "(dv)/(1+v^(2))=(dx)/(x)` On integrating both sides, we get `" "tan^(-1)v=log|x|+C` `rArr" "tan^(-1)((y)/(x))=log|x|+C` |
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