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### Messages - YUTONG ZHANG

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##### Quiz-4 / quiz4 tut5103
« on: October 18, 2019, 02:24:03 PM »
solve the given initial value problem:
y''+4y'+4y=1, y(-1)=2, y'(-1)=1

solution:
r^2+4r+4=0
(r+2)^2=0
r1=r2=-2
Repeated roots
Homo solution is
y(t)=(c1)e^(-2t)+(c2)e^(-2t) (t)
y'(t)=-2(c1) e^(-2t)+(c2) e^(-2t)-2t(c2) e^(-2t)
Plug in initial value y(-1)=2 and y’(-1)=1
c2=5/(e^2)
c1=7/(e^2)

y(t)=(7/e^2)e^(-2t)+(5/e^2)e^(-2t) (t)

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##### Quiz-3 / quiz3 tut5103
« on: October 11, 2019, 02:01:36 PM »
Q: find the wronskian of the given pair of function:x, xe^x

y1(x)=x
y1'(x)=1
y2(x)=xe^x
y2'(x)=e^x+xe^x

w= | x    xe^x
1   e^x+xe^x |

=x(e^x+xe^x )-xe^x=xe^x+(x^2)(e^x)-xe^x
=(x^2)(e^x)

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##### Quiz-2 / tut5103 quiz 2
« on: October 04, 2019, 02:29:18 PM »
(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0, u(x,y)=ye^x

(sin⁡y/y-2e^(-x)  sin⁡x )+((cos⁡y+2e^(-x)  cos⁡x)/y)y'=0
M=sin⁡y/y-2e^(-x)  sin⁡x
dM/dY=〖-y〗^(-2)  sin⁡y+1/y  cos⁡y
N=(cos⁡y+2e^(-x)  cos⁡x)/y
dN/dX=1/y (-2e^(-x)  cos⁡x+2e^(-x) (-sin⁡x ))

Since dM/dY does not equal dN/dX
The equation is not exact.
Multiply μ=ye^x to both side of the equation
Get:

sin⁡〖(y) e^x 〗-2y sin⁡x+(e^x  cos⁡y+2 cos⁡(x))y^'=0
M=sin⁡〖(y) e^x 〗-2y sin⁡x
dM/dY=e^x  cos⁡y-2 sin⁡x
N=e^x  cos⁡y+2 cos⁡x
dN/dX=e^x  cos⁡y-2 sin⁡x

Now dM/dY=dN/dX
The equation is exact.

There exists ϕ(x,y) s.t. dϕ/dx=M, dϕ/dy=N

ϕ=∫M dx=∫〖sin⁡〖(y)〗 e^x 〗-2y sin⁡(x) dx=sin⁡(y) e^x +2y cos⁡(x)+h(y)
dϕ/dy=e^x  cos⁡y+2 cos⁡x+h'(y)
Also, dϕ/dy=N=e^x  cos⁡y+2 cos⁡x
So h’(y)=0

ϕ(x,y)=sin⁡(y) e^x +2y cos⁡(x)=C

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##### Quiz-1 / tut 5103 quiz1
« on: September 27, 2019, 02:58:37 PM »
y^'=(x^2+3y^2)/2xy=x/2y+3y/2x=1/2 (y/x)^(-1)+3/2 (y/x)
v=y/x
v^' x+v=y'
v^' x+v=1/2 v^(-1)+3/2 v
v^' x=1/2 v^(-1)+1/2 v=(1+v^2)/2v
Equation is homogeneous
2v/(1+v^2 ) dv=1/x dx
ln⁡|v^2+1|=ln⁡〖|x|+c〗
v^2+1=cx
y^2/x^2 +1-cx=0
y^2+x^2-〖cx〗^3=0

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