اجوبة الامتحان الشهري الاول للفصل الثاني

حل اسئلة شعبة B
1. 3u_t+ 2u_x =tan?t with u(x,0)=sin?3x
u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?
u(x,t)=g(x-( 2 )/( 3 ) t)+( 1 )/( 3 ) ?_0^t?tan?? d?=sin?(3x-2t)+? ( 1 )/( 3 ) ln?sec?? ?|_0^t
u(x,t)=sin?(3x-2t)+( 1 )/( 3 ) ln?sec?t
An other way dt/(3 )=dx/2=du/tan?t
dt/(3 )=dx/2 ? ( t )/( 3 )+C_1=( x )/( 2 ) ? C=x-( 2 )/( 3 ) t ? C=2C_1
dt/(3 )=du/tan?t ? du=(tan?t dt)/(3 ) ? u(x,t)=( 1 )/( 3 ) ln?sec?t +K
u(x,t)=( 1 )/( 3 ) ln?sec?t +g(x-( 2 )/( 3 ) t) ? u(x,t)=( 1 )/( 3 ) ln?sec?t +sin?(3x-2t)
2. u_t/( 2t )+u_x=u^2 with u(x-1,1)=x
2tdt=dx/1=du/u^2
2tdt=dx/1? x=t^2+C ? C=x-t^2
dx/1=du/u^2 ?dx/1=du/u^2 ? - ( 1 )/( u )=x+K
u(x,t)=(-1)/(x+K)= (-1)/(x+g(x-t^2 ) )
u(x-1,1)=x ? x=(-1)/((x-1)+g(x-1-1) )=(-1)/((x-1)+g(x-2) )
x(x-1)+xg(x-2)=-1 ? g(x-2)=1-x-( 1 )/x
g(x)=1-(x+2)-1/(x+2)
u(x,t)=(-1)/(x+1-(x-t^2+2)-1/(x-t^2+2))
u(x,t)=(t^2-x-2)/(x^2-xt^2+2x+x-t^2+2-(x-t^2+2)^2-1)
u(x,t)=(t^2-x-2)/(x^2-xt^2+3x-t^2+1-(x-t^2+2)^2 )
An other way 2tdt=dx/1=du/u^2
2tdt=dx/1? x=t^2+C ? C=x-t^2
2tdt=du/u^2 ? - ( 1 )/( u )=t^2+K
u(x,t)=(-1)/(t^2+K)= (-1)/(t^2+g(x-t^2 ) )
(x-1,1)=x ? x=(-1)/(1+g(x-1-1) )=(-1)/(1+g(x-2) )
x+xg(x-2)=-1
g(x-2)=(-1-x)/x ? g(x)=(-1-(x+2))/(x+2)=(-1)/(x+2)-1
u(x,t)=(-1)/(t^2-1/((x-t^2 )+2)-1)
3.u_yy (x,y)=sin?(x-2y) with u(x,?)=(3?4) sin?x ,u_y (x,?)=-(1?2) cos?x
u_y (x,y)=(1?2) cos?(x-2y)+f(x)
u_y (x,?)=-(1?2) cos?x ? -(1?2) cos?x=(1?2) cos?(x-2?)+f(x)
-(1?2) cos?x=(2) cos?x+f(x) ? f(x)=-cos?x
u_y (x,y)=(1?2) cos?(x-2y)-cos?x
u(x,y)=-(1?4) sin?(x-2y)-y cos?x+h(x)
u(x,?)=(3?4) sin?x ? (3?4) sin?x=-(1?4) sin?(x-2?)-? cos?x+h(x)
h(x)=sin?x+? cos?x
u(x,y)=-(1?4) sin?(x-2y)-y cos?x+sin?x+? cos?x
4. u_xy (x,y)=3x^2-3y^2- 4xy with u(1,y)=y ,u_x (x,1)=3x^2
u_x (x,y)=3x^2 y-y^3- 2xy^2+f(x)
u_x (x,1)=3x^2 ? 3x^2=3x^2-1-2x +f(x)
f(x)=2x+1
u_x (x,y)=3x^2 y-y^3- 2xy^2+2x+1
u(x,y)=x^3 y-xy^3-x^2 y^2+x^2+x+h(y)
u(1,y)=y ? y=y-y^3-y^2+1+1+h(y)
h(y)=y^3+y^2-2
u(x,y)=x^3 y-xy^3-x^2 y^2+x^2+x+y^3+y^2-2
حل اسئلة شعبة A
1. 2u_t+ 3u_x =cos^2?t with u(x,0)=tan?2x
u(x,t)=g(x-( b )/( a ) t)+( 1 )/( a ) ?_0^t?f(x-( b )/( a ) t+( b )/( a ) ?,?) d?
u(x,t)=g(x-(3 )/( 2 ) t)+( 1 )/( 2 ) ?_0^t?cos^2?? d?
u(x,t)=tan?2(x-(3 )/( 2 ) t)+( 1 )/( 4 ) ?_0^t?(1+cos?2? ) d?
u(x,t)=tan?(2x-3t)+? ( 1 )/( 4 ) (?+( 1 )/( 2 ) sin?2? )?|_0^t
u(x,t)=tan?(2x-3t)+( 1 )/( 4 ) (t+( 1 )/( 2 ) sin?2t )
An other way dt/(2 )=dx/3=du/cos^2?t
dt/(2 )=dx/3 ? ( t )/( 2 )+C_1=( x )/( 3 ) ? C=x-( 3 )/( 2 ) t ? C=3C_1
dt/(2 )=du/cos^2?t ? du=(cos^2?t dt)/(2 ) ? u=( 1 )/( 4 ) ??(1+cos?2t )dt
u(x,t)=( 1 )/( 4 ) (t+( 1 )/( 2 ) sin?2t )+K=( 1 )/( 4 ) (t+( 1 )/( 2 ) sin?2t )+g(x-( 3 )/(2 ) t)
u(x,t)=( 1 )/( 4 ) (t+( 1 )/( 2 ) sin?2t )+tan?(2x-3t)
2. u_t/( 2t )+u_x=u^2 with u(x+1,1)=1
2tdt=dx/1=du/u^2
2tdt=dx/1 ? x=t^2+C ? C=x-t^2
dx/1=du/u^2 ?dx/1=du/u^2 ? - ( 1 )/( u )=x+K
u(x,t)=(-1)/(x+K)= (-1)/(x+g(x-t^2 ) )
u(x+1,1)=1? 1=(-1)/((x+1)+g(x+1-1) )=(-1)/((x+1)+g(x) )
(x+1)+g(x)=-1 ? g(x)=-x-2
u(x,t)=(-1)/(x-(x-t^2 )-2)=(-1)/(t^2-2)
An other way 2tdt=dx/1=du/u^2
2tdt=dx/1? x=t^2+C ? C=x-t^2
2tdt=du/u^2 ? - ( 1 )/( u )=t^2+K
u(x,t)=(-1)/(t^2+K)= (-1)/(t^2+g(x-t^2 ) )
u(x+1,1)=1? 1=(-1)/(1+g(x+1-1) )=(-1)/(1+g(x) ) ? g(x)=-2
u(x,t)=(-1)/(t^2-2)
3.u_yy (x,y)=e^(x-2y) with u(x,0)=( 3 )/( 4 ) e^x ,u_y (x,0)=( 1 )/2 e^x
u_y (x,y)=-( 1 )/( 2 ) e^(x-2y)+f(x)
u_y (x,0)=( 1 )/2 e^x ? ( 1 )/( 2 ) e^x=-( 1 )/( 2 ) e^x+f(x) ? f(x)=e^x
u_y (x,y)=-( 1 )/( 2 ) e^(x-2y)+e^x
u(x,y)=( 1 )/( 4 ) e^(x-2y)+ye^x+h(x)
u(x,0)=( 3 )/( 4 ) e^x ? ( 3 )/( 4 ) e^x=( 1 )/( 4 ) e^x+h(x) ? h(x)=( 1 )/2 e^x
u(x,y)=( 1 )/( 4 ) e^(x-2y)+ye^x+( 1 )/2 e^x
4. u_xy (x,y)=3y^2-3x^2- 4xy with u(x,1)=x ,u_y (1,y)=3y^2
u_y (x,y)=3y^2 x-x^3- 2x^2 y+f(y)
u_y (1,y)=3y^2 ? 3y^2=3y^2-1- 2y+f(y)
f(y)=2y+1
u_y (x,y)=3y^2 x-x^3- 2x^2 y+2y+1
u(x,y)=y^3 x-x^3 y- x^2 y^2+y^2+y+h(x)
u(x,1)=x ? x=x-x^3- x^2+1+1+h(x)
h(x)=x^3+ x^2-2
u(x,y)=y^3 x-x^3 y- x^2 y^2+y^2+y+x^3+ x^2-2