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Topics - maoyafei

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Quiz-5 / Quiz 5
« on: October 31, 2019, 01:08:49 PM »
Question: y''+9y=9sec^2(3t) 0<t<pi/6
y''+9y=0
r^2+9=0
r1=3i
r2=-3i
yc(t)=C1y1(t)+C2y2(t)
=C1cos3t+C2sin3t
W=cos3t*cos3t-sin3t*(-sin3t)=3
u1=-integral of [(sin3t*9sec^2(3t))/3]dt
=-integral of [3sin3t*(1/cos^2(3t)]dt
=-3 integral of (sec3t*tan3t)dt
= - sec3t
=> u1=-sec3t

u2=integral of [(cos3t*9sec^2(3t))/3]dt
=integral of [3 cost3t*(1/cos^2(3t)]dt
= ln/sec3t+tant3t/
=>u2 = ln/sec3t+tan3t/

yp(t)=u1y1+u2y2
=cos3t(-sec3t)+sin3t*(ln/sec3t+tant3t/)
=sin3t(ln/sec3t+tant3t/)-1

y(t)=yc(t)+yp(t)
=C1cos3t+C2sin3t+sin3t(ln/sec3t+tan3t/)-1

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Quiz-3 / TUT 0701 Quiz 3
« on: October 13, 2019, 11:51:38 PM »
Find the general solution of the given differential equation
y''+3y'+2y=0
assume y=e^(rt), y'=re^(rt), y''=(r^2)e^(rt)
r^2+3r+2
(r+1)(r+2)=0
r1=-1, r2=-2

y=C1e^(r1t)+C2e^(r2t)

Ans: y=C1e^(-t) + C2e^(-2t)



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Quiz-2 / TUT 0701 Quiz 2
« on: October 07, 2019, 10:13:08 PM »
Question: (3(x^2)y+2*x*y+y^2) + (x^2+y^2)y' = 0
∂/∂y[𝜇(3(x^2)y+2*x*y+y^3)] = ∂/∂x[𝜇(x^2+y^2)]
∂𝜇/∂y (3(x^2)y+2*x*y+y^3) + ∂𝜇/∂x (x^2+y^2) + 𝜇(2*x)
∂𝜇/∂y (3(x^2)y+2*x*y+y^3) + 𝜇(3*x^2+3*y^2) = ∂𝜇/∂x (x^2+ y^2)
∂𝜇/∂y = 0      ∂𝜇/∂x = d𝜇/dx
𝜇(3*x^2+3*y^2) = d𝜇/dx (x^2+y^2)
(Divide both sides by (x^2+y^2)
3𝜇=d𝜇/dx
\int 1/𝜇d𝜇 = \int 3dx
ln(𝜇) = 3*x
𝜇=e^(3*x)

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Quiz-1 / TUT 0701 Quiz 1
« on: September 27, 2019, 03:07:41 PM »
Please see the attachment:

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