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The solution of the differential equation `(xy^4 + y) dx-x dy = 0,` isA. `4x^(4)y^(3)+3x^(3)=Cy^(3)`B. `3x^(3)y^(4)+4y^(3)=Cx^(3)`C. `3x^(4)y^(3)+4x^(3)=Cy^(3)`D. none of these |
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Answer» Correct Answer - C We have, `(xy^(4)+y)dx-xdy=0` `rArr" "(dy)/(dx)=(xy^(4)+y)/(x)` `rArr" "(dy)/(dx)-(y)/(x)=y^(4)` `rArr" "(1)/(y^(4))(dy)/(dx)+((-1)/(y^(3)))(1)/(x)=1` Let `-y^(-3)=v." Then, "3y^(-4)(dy)/(dx)=(dv)/(dx)` `therefore" "(1)/(3)(dv)/(dx)+(v)/(x)=1rArr (dv)/(dx)+(3)/(x)v=3" ...(i)"` This is a linear differential equaiton with integrating factor `x^(3)`. Multiplying both sides of (i) by `x^(3)` and integrating, we get `vx^(3)=(3x^(4))/(4)+C` `rArr" "(-x^(3))/(y^(3))=(3x^(4))/(4)+C` `rArr" "-4x^(3)=3x^(4)y^(3)+4y^(3)C` `rArr" "3x^(4)y^(3)+4x^(3)=-4Cy^(3)rArr 3x^(4)y^(3)+4x^(3)=lambday^(3)` |
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