Answer to Question #199326 in Differential Equations for Anushka Bhasme

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

Solutions:-

1.

$P(x,y)=x^2-4xy-2y^2$

$Q(x,y)=y^2--4xy-2x^2$

$\frac{\partial P(x,y)}{\partial y}=\frac{\partial Q(x,y)}{\partial x}=-4x-4y$

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

$f(x,y)=\int(x^2-4xy-2y^2)dx=x^3/3-2x^2y-2xy^2+g(y)$

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(x^3/3-2x^2y-2xy^2+g(y))=-2x^2-4xy+\frac{dg(y)}{dy}$

$-2x^2-4xy+\frac{dg(y)}{dy}=-2x^2-4xy+y^2$

$\frac{dg(y)}{dy}=y^2$

$g(y)=y^3/3$

$f(x,y)=x^3/3-2x^2y-2xy^2+y^3/3$

The solution is $f(x,y)=c$

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

2.

$P(x,y)=y+2xy-tany$

$Q(x,y)=x^2--sec^2y-xtan^2y$

$\frac{\partial P(x,y)}{\partial y}=\frac{\partial Q(x,y)}{\partial x}=2x-sec^2y+1$

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

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

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(xy+x^2y-xtany+g(y))=x+x^2-xsec^2y+\frac{dg(y)}{dy}$

$x+x^2-xsec^2y+\frac{dg(y)}{dy}=x^2--sec^2y-xtan^2y$

$\frac{dg(y)}{dy}=-sec^2y$

$g(y)=-tany$

$f(x,y)=xy+x^2y-tany-xtany$

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

5.

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

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

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

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

3.

$Mdx+Ndy=0$

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

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

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

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

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

Integrating factor:

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

Multiplying by the I.f. we get:

$(2xe^y+2x/y+1/y^3)dx+(x^2e^y-x^2y^2-3x/y^4)dy=0$

which is exact.

Then:

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

Solution:

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

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:

$ylogx-xsiny=c$