- #1
krusty the clown
I am supposed to make this equation exact and then solve it. I can do the problem, but I don't understand why I am doing what I am. Is the final answer right?
[tex] cos(x) dx + (1 + \frac {2}{y}) sin(x) dy = 0 [/tex]
[tex] M(x,y)=cos(x) [/tex]
[tex]N(x,y)=(1 + \frac{2}{y}) sin(x) [/tex]
[tex] \frac{\partial M(x,y)}{\partial y} = 0[/tex]
[tex] \frac{\partial N(x,y)}{\partial x} = (1 + \frac {2}{y}) cos(x)[/tex]
The integration factor is then
[tex] u(x)= e^{\int\frac{(1+\frac{2}{y})cos(x)}{cos(x)}dy)}=e^{y+2\ln(y)} = y^2 e^y[/tex]
Multiplying the integration factor by both sides,
[tex] y^2 e^y cos(x) dx + (y^2 e^y + 2ye^y)sin(x)dy=0 [/tex]
I confirmed that this was exact, the new M(x,y) and N(x,y) are
[tex] M(x,y)=y^{2}e^{y}cos(x) \ \ \ N(x,y)=ye^{y}(y+2)sin(x)[/tex]
[tex] g(x,y)=\int{M(x,y)dx}+g(y)=\int{(2ye^{y}cos(x)+y^{2}e^{y}cos(x))dx}+g(y)=sin(x)2ye^{y}+sin(x)y^{2}e^{y}+g(y) [/tex]
[tex]N(x,y)=\frac{\partial}{\partial y}(2ye^{y}sin(x)+y^{2}e^{y}sin(x)+g(y))[/tex]
[tex]=e^{y}sin(x)(2+4y+y^{2})+g'(y) [/tex]
set this equal to N(x,y), and solve for g'(x)
[tex]e^{y}sin(x)(2+4y+y^{2})+g'(y)=y^{2}e^{y}sin(x)+2ye^{y}sin(x)[/tex]
[tex]g'(x)=-4e^{y}sin(x) \ \ \ \ \ \ \ \ \ \\ \ g(x)=-4e^{y}sin(x)+c [/tex]
[tex]g(x,y)=e^{y}sin(x)(2y+y^{2}-4)=C[/tex]
Thanks-Erik
[tex] cos(x) dx + (1 + \frac {2}{y}) sin(x) dy = 0 [/tex]
[tex] M(x,y)=cos(x) [/tex]
[tex]N(x,y)=(1 + \frac{2}{y}) sin(x) [/tex]
[tex] \frac{\partial M(x,y)}{\partial y} = 0[/tex]
[tex] \frac{\partial N(x,y)}{\partial x} = (1 + \frac {2}{y}) cos(x)[/tex]
The integration factor is then
[tex] u(x)= e^{\int\frac{(1+\frac{2}{y})cos(x)}{cos(x)}dy)}=e^{y+2\ln(y)} = y^2 e^y[/tex]
Multiplying the integration factor by both sides,
[tex] y^2 e^y cos(x) dx + (y^2 e^y + 2ye^y)sin(x)dy=0 [/tex]
I confirmed that this was exact, the new M(x,y) and N(x,y) are
[tex] M(x,y)=y^{2}e^{y}cos(x) \ \ \ N(x,y)=ye^{y}(y+2)sin(x)[/tex]
[tex] g(x,y)=\int{M(x,y)dx}+g(y)=\int{(2ye^{y}cos(x)+y^{2}e^{y}cos(x))dx}+g(y)=sin(x)2ye^{y}+sin(x)y^{2}e^{y}+g(y) [/tex]
[tex]N(x,y)=\frac{\partial}{\partial y}(2ye^{y}sin(x)+y^{2}e^{y}sin(x)+g(y))[/tex]
[tex]=e^{y}sin(x)(2+4y+y^{2})+g'(y) [/tex]
set this equal to N(x,y), and solve for g'(x)
[tex]e^{y}sin(x)(2+4y+y^{2})+g'(y)=y^{2}e^{y}sin(x)+2ye^{y}sin(x)[/tex]
[tex]g'(x)=-4e^{y}sin(x) \ \ \ \ \ \ \ \ \ \\ \ g(x)=-4e^{y}sin(x)+c [/tex]
[tex]g(x,y)=e^{y}sin(x)(2y+y^{2}-4)=C[/tex]
Thanks-Erik