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Ask QuestionPosted by Ashish Varma 4 years, 5 months ago
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I guess you want to find the Particular Solution of the equation {tex}x(x^2-1) {dx \over dy} = 1 \ at\ y =0, x=2{/tex}
{tex}It \ can \ be\ written \ as => x(x^2-1) dx = dy{/tex}
=> {tex}(x^3-x) dx =dy{/tex}
On Integrating both side
{tex}{x^4 \over 4 }- {x^2 \over 2} = y{/tex} + c ....................1
Put x=2 & y=0 in equation 1
c = 2
Now put value of c in 1
{tex}{x^4 \over 4 }- {x^2 \over 2} = y + 2{/tex}
This is the Particular Solution of given differential equation.
Equation 1 is the General Solution of given differential equation.
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Shekhar Yadav 4 years, 4 months ago
0Thank You