Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : 2015

$(x^3y^3 - xy) = dx /dy\\$

$\dfrac {dx}{dy}+xy=x^3y^3\\\ $

Dividing by $x^3$ we get $\\ $

$\dfrac 1{x^3}\dfrac {dx}{dy}+\dfrac 1{x^2}y=y^3----- \gt(1)\\\ $

Let $\dfrac1{x^2}=v\hspace {1cm}\text{diff w.r.t y}\\\ $

$\therefore\dfrac {dv}{dy} =\dfrac{-2}{x^3}\dfrac {dx}{dy}\\\ $

Substituting these values in eq (1) $\\ $

We get $\\ $

$\dfrac {-1}2\dfrac {dv}{dy}+ vy=y^3 \\ $

$\therefore \dfrac {dv}{dy}+(-2y)v=-2y^3 \\ $

Which is of the type $\\ $

$\dfrac {dv}{dy}=P'v=Q' \\ $

Where $p’ = -2y$ and $Q’ = -2y^3 \\ $

Therefore it is a linear differential equation $\\ $

$\therefore If=e^{\int fdy}=e^{\int -2ydy} \\ $

$=e^{\dfrac{-2y^2}2} \\ $

$=e^{-y^2} \\ $

Solution is $ \\ $

$v\times e^{-y^2}=\int-UC^UdU \\ $

Integration by parts we get $\\ $

$=-\Bigg[U\int e^UdU-\int\Bigg[\dfrac {dU}{dU}\int C^UdU\Bigg]dU\Bigg] \\ $

$=-[Ue^U-e^U]+C \\ $

$=e^U(1-U)+C \\ $

Resubstituting values of U we get $\\ $

$V\times e^{-y^2}=e^{-y^2}(1+y^2)+C \\ $

Where $V=\dfrac 1{x^2}\\ $

$\therefore \dfrac 1{x^2}=1+y^2+Ce^{y^2}\\ $