InterviewSolution
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InterviewSolution
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Solve the following differential equations : (1) ` (dy)/(dx) = (y + sqrtx^(2)+y^(2))/x ` |
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Answer» `(dy)/(dx)=(y+sqrt(x^(2)y^(2)))/(x)" …(i)"` Putting `y=vx rArr (dy)/(dx)=v+x(dv)/(dx)` `therefore" "v=(y)/(x)` `therefore" "` Equation (i) becomes, `v+x(dv)/(dx)=(vx+sqrt(x^(2)+v^(2)x^(2))/(x))` `therefore" "v+x(dv)/(dx)=v+sqrt(1+v^(2))` `rArr" "x(dv)/(dx)=sqrt(1+v^(2))` `rArr" "(1)/(sqrt(1+v^(2)))dv=(1)/(x)dx` On integration, we get `int(1)/(sqrt(1+v^(2)))dv=int(1)/(x)dx` `rArr" "log lv+sqrt(1+v^(2))|=log|x|+logx` `rArr" "log|(y)/(x)+sqrt(1+(y^(2))/(x^(2)))|=log|cx|` `rArr " "y+sqrt(x^(2)+y^(2))=cx^(2)` This is the general solution. |
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