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  • Answer please! - Ordinary Differential Equations - BITSAT-2

The solution of differential equation (2x-10y^{3})\frac{dy}{dx}+y=0 is

  • Option 1)

    xy^{2}=y^{5}+C

  • Option 2)

    xy^{2}+2y^{5}=C

  • Option 3)

    xy^{2}=2y^{5}+C

  • Option 4)

    None of these

 

Answers (1)

use the concept of

Linear Differential Equation -

\frac{dy}{dx}+Py= Q

- wherein

P, Q are functions of x alone.

 

 \left ( 2x-10y^{3} \right )\frac{dy}{dx}+y=0

 =>2x-10y^{3}=-y\frac{dx}{dy}

=>-\frac{2x}{y}-10y^{2}=\frac{dx}{dy}

=>\frac{dx}{dy}+\frac{2}{y}x=10y^{2}

p=\frac{2}{y}

\therefore 2\int \frac{dy}{y}=2logy=logy^{2}

\therefore Solution\: is\: x.y^{2}=\int 10y^{5}dy

xy^{2}=\frac{10y^{5}}{5}+C

\therefore xy^{2}=2y^{5}+C


Option 1)

xy^{2}=y^{5}+C

Option is incorrect

Option 2)

xy^{2}+2y^{5}=C

Option is incorrect

Option 3)

xy^{2}=2y^{5}+C

Option is correct

Option 4)

None of these

Option is incorrect

Posted by

Vakul

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