InterviewSolution
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InterviewSolution
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The general solution of the differential equation `(y^(2)-x^(3))dx-xydy=0 (x ne0)` is (where, C is a constant of integration)(A) `y^(2)-2x^(2)+Cx^(3) =0`(B) `y^(2)+2x^(3)+Cx^(2) =0`(C) `y^(2)+2x^(2)+Cx^(3) =0`(D) `y^(2)-2x^(3)+Cx^(2) =0`A. (a) `y^(2)-2x^(2)+Cx^(3) =0`B. (b) `y^(2)+2x^(3)+Cx^(2) =0`C. (c) `y^(2)+2x^(2)+Cx^(3) =0`D. (d) `y^(2)-2x^(3)+Cx^(2) =0` |
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Answer» Correct Answer - (b) Given differential equation is `(y^(2)-x^(3))dx-xy dy=0, (xne0)` `rArr xydy/dx-y^(2)=-x^(3)` Now, put `y^(2)=trArr 2y dy/dx = (dt)/dx rArr ydy/dx=1/2 (dt)/dx` `therefore x/2 (dt)/dx-t=-x^(3)` `rArr (dt)/dx -2/xt =-2x^(2)` which is the linear differential equation of the form `(dt)/dx+Pt=Q.` Here, `p=-2/xand Q=-2x^(2).` Now, `IF=e^(-f2/x dx)=1/x^(2)` `therefore` solution of the linear differential equation is `(IF) t=intQ(IF)dx+ lambda` [where `lambda` is integrating constant ] `therefore t(1/x^(2))=-2int(x^(2) xx 1/x^(2))dx + lambda` `rArr t/x^(2) =-2x+lambda` `rArr y^(2) /x^(2) +2x =lambda` `[therefore t=y^(2)]` `rArr y^(2) + 2x^(3) - lambda x^(2) = 0` or `y^(2) +2x^(3) +cx^(2)=0` " [let C"=-`lambda]` |
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