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# X(X^2-1)dx/dy=1:y=0 x=2

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X(X^2-1)dx/dy=1:y=0 x=2
• 2 answers

Shekhar Yadav 2 years, 8 months ago

Video

Arpit Airen 2 years, 10 months ago

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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