Answer to Question #200137 in Differential Equations for Anuj

Question

1. (x²-4xy-2y²)dx + (y²-4xy-2x²)dy=0
2. (x²-xtan²y+sec²y)dy = (tany-2xy-y)dx
3. (2xy⁴e^y+2xy³+y)dx + (x²y⁴e^y-x²y²-3x)dy=0
4. (y/x secy-tany)dx+(secy logx-x)dy=0
5. y dy/dx +4x/3 -y²/3x =0

Answer 1

1.

$M=x^2-4xy-2y^2$

$N=y^2-4xy-2x^2$

$\partial M/\partial y=-4x-4y=\partial N/\partial x$

so, equation is exact

$\int Mdx=\int (x^2-4xy-2y^2)dx=x^3/3-2x^2y-2y^2x$

$\int Ndy=\int (y^2-4xy-2x^2)dx=y^3/3-2x^2y-2y^2x$

omitting -2xy² - 2x²y which already occur in ∫Mdx:

$x^3/3-2x^2y-2y^2x+y^3/3=c$

2.

$M= -tany+2xy+y$

$N=x^2-xtan^2y+sec^2y$

$\partial M/\partial y=2x-tan^2y=\partial N/\partial x$

so, equation is exact

$F=\int Mdx=\int ( -tany+2xy+y)dx=x^2y+xy-xtany+g(y)$

$\partial F/\partial y=x^2+x-xsec^2y+g'(y)$

$x^2+x-xsec^2y+g'(y)=N=x^2-xtan^2y+sec^2y$

$g'(y)=sec^2y$

$g(y)=\int sec^2y dy=tany +c$

$F=x^2y+xy-xtany+tany+c=0$

3.

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

$N=x^2y^4e^y-x^2y^2-3x$

$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}=\frac{-8xy^2-4-8xy^3e^y}{2xy^4e^y+2xy^3+y}=-\frac{4}{y}$

Integrating factor:

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

then:

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

$M_1=2xe^y+2x/y+1/y^3$

$N_1=x^2e^y-x^2/y^2-3x/y^4$

$\int M_1dx+\intop$(terms of N₁ free from x)$dy=c$

$\int (2xe^y+2x/y+1/y^3)dx=c$

$x^2e^y+x^2/y+x/y^3=c$

4.

$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}=-tany$

I.f.:

$e^{\int-tanydy}=cosy$

Then:

$(y/x secy-tany)cosydx+(secy logx-x)cosydy=0$

$(y/x-sinx)dx-(xcosy-logx)dy=0$

$\int (y/x-sinx)dx=ylogx-xsiny$

Solution:

$ylogx-xsiny=c$

5.

$\frac{3yy'}{y^2-4}=x$

$\int\frac{3ydy}{y^2-4}=\int xdx$

$3ln(y^2-4)=x^2+c_1$

$y=\sqrt{ce^{x^2/3}+4}$