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Messages - Sally

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1
Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 08:27:05 AM »
(-ysin(x)+(y^3)cos(x))+(3cos(x)+5(y^2)sin(x))y ’=0
a). My=-sin(x) +(3y^2)(cos(x)) Nx = - 3s in ( x ) + 5( y ^ 2) ( c o s( x ) )
R 2= [ My - N x ] /M = [ - si n ( x ) +( 3 y ^2 ) co s ( x ) + 3s in ( x ) - 5( y ^ 2) c o s( x ) ] / [ - y si n ( x ) +( y ^ 3) ( c o s( x ) ) = - 2 / y
u=e^(- ∫(2/y)dy)=y^2 u(-ysin(x)+(y^3)cos(x))+u(3cos(x)+5(y^2)sin(x))y ’=0
Ø=∫u(-ysin(x)+(y^3)cos(x)) dx =(y^3)cos(x)+(y^5)sin(x)
Ø y =3( y ^ 2) c o s( x ) + 5( y ^ 4) s i n ( x )
=3(y^2)cos(x)+5(y^4)sin(x)+h ’(y) h’(y)=0
h(y)=c C=(y^3)cos(x)+(y^5)sin(x)

b) x=π/4 y=√2
C=2+4=6

2
Quiz-4 / TUT 0501
« on: October 18, 2019, 02:15:33 PM »
y”+2y’+2y=0.  y(π/4)=2.      y’(π/4)=-2。        (note: π is “pie”)

r^2+2r+2=0
r1=-1+i
r2=-1-i

y=C1e^(-t)cos(t)+C2e^(-t)sin(t)

y(π/4)=2=[e^(-π/4)](1/√2)(C1+C2)
C1+C2=2/([e^(-π/4)] (1/√2))

y’(t)=-C1[e^(-t)]cos(t)-C1[e^(-t)]sin(t)
-2=-2[e^(-π/4)](1/√2)C1
C1=√2/[e^(π/4)]=C2
y=(√2/[e^(π/4)])*e^(-t)cos(t)+(√2/[e^(π/4)])*e^(-t)sin(t)

3
Quiz-3 / Tut 0501
« on: October 11, 2019, 01:53:59 PM »
(t^2)y”-t(t+2)y’+(t+2)y=0
y”+(-1)(t)(t+2)(y’)/(t^2)+(t+2)(y)/(t^2)
P(t)=-(t+2)/t=-(1+(2/t))
W=ce^(-∫P(t)dt) =ce^(∫1+(2/t)dt) =c(e^t)*(e^(ln(t^2)) =c(t^2)(e^t)

4
Quiz-1 / TUT0501
« on: September 27, 2019, 10:07:39 AM »
Q:   ty’ +(t+1)y=t. y(ln(2))=1
Ans:
y’+[(t+1)/t]y=1
p(t)=[(t+1)/t]
u=e^(∫[(t+1)/t]dt)=te^t
te^t*y’+[(t+1)/t]*te^t*y=te^t
te^t *y=∫te^tdt
      u=t   du=1dt
      v=e^t   dv=e^tdt
te^t *y = te^t-∫e^t dt = te^t - e^t +c
y=1-(1/t)+(c/(te^t))

y=1 t=ln2
1=1-1/ln2 +c/(ln2)
C=2
y=1-1/t+2/(te^t)

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