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Messages - chaoy

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1
Quiz-2 / TUT0202Quiz2
« on: October 04, 2019, 02:24:05 PM »
(3x^2y+2xy+y^3)+(x^2+y^2){y}'=0
\because My=3x^2+2x+3y^2,
         Nx=2x,
\therefore My\neq Nx,given DE is not exact;
we need to find a function \mu (x,y)s.t. the equation \mu (3x^2y+2xy+y^3)+\mu (x^2+y^2){y}'=0 is exact;
d(\mu (3x^2y+2xy+y^3))/dx = d(\mu (x^2+y^2))/dx;
(d\mu/dy) (3x^2y+2xy+y^3)+ \mu (3x^2+2x+3y^2)=(d\mu/dx)(x^2+y^2)+\mu (2x);
(d\mu/dy) (3x^2y+2xy+y^3)+ \mu (3x^2+3y^2)=(d\mu/dx)(x^2+y^2);
suppose \mu is a function of x only;
then d\mu /dy=0;
then \mu (3x^2+3y^2)=(d\mu/dx)(x^2+y^2)
divide both sides by (x^2+y^2);
3\mu =d\mu /dx
which is a seperable equation;
\int (1/\mu)d\mu=\int 3dx;
ln(\mu)=3x;
\mu=e^(3x);
thus, \mu =e^(3x)is an integration factor for the given DE;
multiply by \mu =e^(3x);
e^(3x)(3x^2y+2xy+y^3)+ e^3x(x^2+y^2){y}'=0;
which is exact;
\phi (x,y)=C;
d\phi /dx=e^(3x)(3x^2y+2xy+y^3)and d\phi /dy=e^(3x)(x^2+y^2);
\phi (x,y)=e^(3x)(x^y+(1/3)y^3)+ g(x);
d\phi /dx=3e^(3x)(x^y+(1/3)y^3)+e^(3x)(2xy)+g({x})';
d\phi /dx=e^(3x)(3x^2y+2xy+y^3)
g({x})'=0;
g(x)=C;
\phi (x,y)=e^(3x)(x^2y+(1/3)y^3)=C;
the general solution:e^(3x)(x^2y+(1/3)y^3)=C

2
Quiz-1 / TUT0202Quiz1
« on: September 27, 2019, 06:31:12 PM »
(y)'-y=1+3sin(t),y(0)=y0
 
p(t)=-1,q(t)=1+3sin(t)
\mu (t)=e^{\int p(t)dt}=e^{\int (-1)dt}=e^{-t}
\mu (y)'-e^{-t}y=e^{-t}(1+3sin(t))
d(e^{-t}y)/dt=e^{-t}+3e^{-t}sin(t)
\int(d(e^{-t}y)/dt)=\int(e^{-t}+3e^{-t}sin(t)dt)
e^{-t}y=-e^{-t}+(-3/2)sin(t)e^{-t}+(-3/2)cos(t)e^{-t}+c
y(t)=-1+(-3/2)sin(t)+(-3/2)cos(t)+ce^{t}
y(0)=y0=-1+(-3/2)sin(0)+(-3/2)cos(0)+ce^{0}=-1+(-3/2)+c
c=y0+(5/2)
y(t)=-1+(-3/2)sin(t)+(-3/2)cos(t)+(y0+(5/2))e^{t}
if y0<-5/2,y0+(5/2)<0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(y0+(5/2))e^{-t}=-\infty
if y0>-5/2,y0+(5/2)>0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(y0+(5/2))e^{-t}=+\infty
if y0=-5/2,y0+(5/2)=0
   \lim_{t\to \infty}y(t)=\lim_{t\to \infty}(-1+(-3/2)sin(t)+(-3/2)cos(t))is finite
so y0=-5/2

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