InterviewSolution
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InterviewSolution
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The differential equations, find a particular solution satisfying the given condition: `(1+x^2)(dy)/(dx)+2x y=1/(1+x^2); y=0`when `x = 1` |
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Answer» `(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2))` `implies (dy)/(dx)+(2x)/(1+x^(2))y=(1)/((1+x^(2))^(2))` Here, `P=(2x)/(1+x^(2))` and `Q=(1)/((1+x^(2))^(2))` `I.F.=e^(intPdx)=e^(int(2x)/(1+x^(2))dx)` `:. =e^(log(1+x^(2))=(1+x^(2))` and general solution : `y*(1+x^(2))=int(1)/((1+x^(2))^(2))(1+x^(2))dx+c` `=int(1)/(1+x^(2))dx+c` `y(1+x^(2))=tan^(-1)x+c`.........`(1)` Given, `y=0` at `x=1` `0=tan^(-1)1+c` `impliesc=-(pi)/(4)` Therefore, from equation `(1)`, particular solution is `y(1+x^(2))=tan^(-1)x-(pi)/(4)` |
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