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Differential equation Solve $(2xy^4e^y + 2xy^3+ y)dx + (x^2y^4e^y- x^2y^2- 3x) dy = 0$

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

Marks : 6

Year : 2013

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We have equation as $Mdx + Ndy = 0$

$M = 2xy^4e^y + 2xy^3 + ys$

$N = x^2y^4e^y – x^2y^2 – 3x$

$\dfrac {\partial M}{\partial y}=2x(y^4e^y+4y^3e^y)+6xy^2+1$

$\dfrac {\partial N}{\partial y}=2xy^4e^y -2xy^2-3$

$\dfrac {\dfrac {\partial M}{\partial y}-\dfrac {\partial N}{\partial y}}M=\dfrac {2xy^4e^y-2xy^2-3-2xy^4e^y-8xy^3e^y-6xy^2-1}{2xy^4e^y+2xy^3+y}$

$=\dfrac {-8xy^2-4-8xy^3e^y}{2xy^4e^y+2xy^3+y}$

$=\dfrac {-4}y\Bigg[\dfrac{2xy^2+1+2xy^3e^y}{2xy^2+1+2xy^3e^y}\Bigg]$

$=\dfrac {-4}y f(y)$

If $e^{\int -(4/y)dy}=e^{-4\log y}=e^{\log (1/y^4)}=1/y^4$

Multiplying by the If we get

$\Bigg(2xe^y+\dfrac {2x}y+\dfrac 1{y^3}\Bigg)dx +\Bigg(x^2e^y-\dfrac {x^2}{y^2}-\dfrac {3x}{y^4}\Bigg)dy=0$

Which is exact.

$\int Mdx=\int(2xe^y+\dfrac {2x}y+\dfrac 1{y^3})dx$

$=x^2e^y+\dfrac {x^2}y+\dfrac x{y^3}$

$\int $ (Terms in N free from x)dy =0

$\therefore$ The complete solution is $x^2e^y +\dfrac {x^2}y+\dfrac x{y^3}=c$

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