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Messages - Michael Zhang

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Quiz-3 / TUT0302 QUIZ1
« on: October 11, 2019, 03:22:53 PM »
Q: Verify that the functions y_1 and y_2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
y”+4y=0
y_1(t)=cos(2t), y_2(t)=sin(2t)

y_1’(t)=-2sin(2t)
y_1”(t)=-4cos(2t)
y_2’(t)=2cos(2t)
y_2”(t)=-4sin(2t)
Plug y_1 into the given equation
-4cos(2t)+4cos(2t)=0
Plug y_2 into the given equation
-4sin(2t)+4sin(2t)=0

So, y_1 and y_2 are solutions of given equation

W= det(y_1,y_2,y_1',y_2') = y_1(t)y_2’(t) – y_2(t)y_1’(t)
=cos(2t)2cos(2t) – sin(2t)(-2)sin(2t)
=2cos^2(2t)+2sin^2(2t)
2[cos^2(2t)+sin^2(2t)]
=2
Since W≠0, it is a fundamental set of solutions

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Quiz-2 / TUT0302 QUIZ2
« on: October 04, 2019, 02:39:42 PM »
Q: 1+((x/y)-siny)y’= 0

Let M(x,y)=1 , N(x,y)= ((x/y)-siny)
Then, ∂/∂y{M(x,y)}=0 , ∂/∂x{N(x,y)}=1/y
Notice that
(N_x – M_y)/M = 1/y
It contains y only, so
dμ/dy = [(N_x – M_y)/M] μ = μ/y
μ=y
Multiplying original equation by μ(y),we get
y + (x-ysin(y))y’= 0
Now,this equation is exact, since
M_y = N_x
Therefore,
∃φ(x,y) s.t. φ_x = M = y
φ=∫ydx= xy + h(y)
φ_y= x +h’(y)
Also, φ_y=N=x-ysin(y)
So h’(y)=ysin(y)
h(y)= ∫ysin(y) dy=ycos(y)-sin(y)+c
Thus
φ=xy+ ycos(y)-sin(y)=C

3
Quiz-1 / TUT0302 QUIZ1
« on: September 27, 2019, 07:44:53 PM »
Q:dy/dx = (x^2 - 3y^2)/(2xy)

Both sides divide x^2,
dy/dx = [1-3(y/x)^2]/[2(y/x)]
Let u=y/x,then,y=ux
dy/dx = d(ux)/dx
(du/dx)x + u = (1-3u^2)/2u
(du/dx)x = [(1-3u^2)/2u] - u = (1-5u^2)/2u
[2u/(1-5u^2)]du = (1/x)dx
∫[2u/(1-5u^2)]du=∫(1/x)dx
LHS uses substitution,
t=1-5u^2
dt= (-10u)du
=(-1/5)∫(1/t)dt
So,LHS= (-1/5)In(1-5u^2)
RHS= ln(x)+c
In(1-5u^2) = -5In(x)-5c
1-5(y/x)^2 = e^(-5In(x)-5c)
5(y/x)^2 = 1 - [e^(-5In(x))]/e^5c]= 1-  [x^(-5)]/[e^5c] = (e^5c-1)/[(e^5c)x^5]
(y/x)^2 = (e^5c-1)/5[(e^5c)x^5]
y^2 = [(e^5c-1)(x^2)]/5[(e^5c)x^5]
y = ±√[(e^5c-1)(x^2)]/5[(e^5c)x^5]

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