solve question 5 and question 6
Answers
5 a) we take the complementary D.E. part of the given ODE.
y’’ – 2 tan x y’ + 5 y = 0
This is of the form: y’’+ p(x) y’ + q(x) y = 0
p(x) = -2 tan x q(x) = 5
Solution: y(x) = u(x) * v(x).
Integrating factor= v(x) = exp(- integral p(x)/2 dx)
v = exp(int tanx dx) = exp(Ln sec x) = sec x
v’ = sec x tan x , v’’ = sec x tan^2 x + sec^3 x= secx (2tan^2 x +1)
Q(x) = v’’ + p v’+ q v = sec x [2 tan^2 x + 1 – 2 tan^2 x + 5]= 6 sec x
Normal form of ODE: v(x) u’’(x) + Q(x) u(x) = 0
u’’(x) + 6 u(x) = 0 => ω^2 = 6
u(x) = A Sin (ωx+θ)
So complementary solution: y(x) = A Sin(ωx+θ) Sec x
Particular Solution of: y’’ – 2 tan x y’ + 5 y = e^x sec x
y(x) = K e^x Sec x, y’= K e^x Sec x (tan x + 1)
y’’(x) = 2 K e^x Sec x [tan^2 x + tan x + 1]
So 2K e^x Sec x [tan^2 x +tan x+1]- 2K e^x sec x tan x (tan x+1) + 5K e^x sec x = e^x sec x
2K [tan^2 x +tan x+1]- 2K tan x (tan x+1) + 5K = 1
K = 1/7
Hence the complete solution: A Sin(√6 x+θ) Sec x + 1/7*e^x Sec x
===========================
5 b) Assuming a LCR series circuit.
- R i(t) – L di(t)/dt + Q/C = E(t)=0, i(t) = -dQ(t)/dt, Q(t)=charge on Capacitor.
d^2 Q/dt^2 + R/L dQ/dt + Q(t)/LC = 0…….ODE : p(t)= R/L , q(t)=1/LC
Q(t)= u(t) v(t). v(t) = exp(- integral R/2L dt) = exp(-R t/2L)
v’ = -R/2L * v, v’’ = R^2/4L^2 * v
Complementary solution : v u’’ + (v’’ + p v’ + q v) u = 0
=> v u’’+ (R^2/4L^2 – R^2/2L^2 + 1/LC) v u = 0
=> u’’ + (1/LC - R^2/4L^2) u = 0 => Sinusoidal solution
=> ω = √[4LC – R^2C^2] /2LC]=40 rad/sec, u(t) = A Sin(ωt + θ)
Q(t) = A exp(-Rt/2L) Sin(ωt+θ) = A exp(-20 t) Sin(40t+θ)
Q(t=0) = 5 Coul. => A Sinθ = 5
i(t)=-Q’(t) = - A exp(-20t)*[40 Cos(40t+θ)- 20 Sin(40t+θ)]
i(t=0) = 0, So Tanθ =2, θ = 1.107 rad or 63.4 deg.
So A = 5 /Sinθ = 5.59 Coul
Q(t=0.01sec) = 5.59 exp(-0.2) Sin(0.4+1.107) = 4.56 Coul
====================
5c) x y’ + y Ln y = x y e^x
Let y = e^z or Ln y = z and y’(x) = e^z * z’(x)
=> ODE : x z’ e^z + e^z z = x e^z e^x
z’ + 1/x * z = e^x
Compare with z’(x) + p(x) z(x) = q(x). p(x)=1/x , q(x)=e^x
Integrator fn = μ(x)=exp( int p(x) dx] = exp(Ln x) = x
Solution: z(x) = 1/μ(x) * [integral μ(x) q(x) dx + C]
= 1/x * [x e^x – e^x + C]
y(x) = exp[ e^x – e^x /x + C/x]
=================================
6 ) b) d^2 x/dt^2 + 2 m dx/dt + n^2 x = a cos pt
C.F. : x’’ + 2 m x’ + n^2 x = 0
x(t) = u(t) v(t) , p(t)= 2m q(t)= n^2
v(t) = exp[- integral m dt] = exp(-mt),
v ’ = - m v, v’’ = m^2 v
Q(t) = v’’+ p v’ + q v = (m^2 - 2m^2 + n^2) v
N. F. : v(t) u’’(t) + Q(t) u(t) = 0 or u’’ + (n^2-m^2) u =0
As n > m >0, ω = √(n^2-m^2)
u(t) = A Sin(ωt+θ) =>
Complementary Solution: x(t) = A exp(-mt) Sin(ωt+θ)
So after about t = 6/m the value of x(t)=amplitude is very low.
So the vibrations are damped out down.
Particular solution:
x2(t)= A Sin (pt+ф) + B
x2’(t) = A p Cos(pt+ф) , x2’’(t) = -A p^2 Sin(pt+ф)
=> -A p^2 Sin(pt+ф) + 2m A p Cos(pt+ф) + n^2 A Sin(pt+ф)+n^2B = a cos pt
A (n^2-p^2) sin pt cosф + A(n^2-p^2) cos pt sinф + 2mAp Cos pt cosф
– 2 mAp sin pt sinф + n^2 B = a cos pt
Compare coefficients of sin pt and cos pt...
A (n^2-p^2) cosф -2mAp sinф = 0 => Tan ф = (n^2 – p^2)/2pm)
A (n^2-p^2) sinф +2mAp cosф = a => 2Apm [sin^2ф+cos^2ф]cosф = a
A = a Sec ф/(2pm)
Complete Solution: A exp(-mt) Sin(ωt+θ) + a Sec ф /(2pm) * Sin(pt+ф)