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Differential equations - 2nd order nonhomogenous eq'n

  • #1
162
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differential equations - 2nd order homogenous eq'n

sorry the title should read 2nd order homogenous eq'n, not nonhomogenous

Find the general solution of the equation:
(1+t^2)d^2y/dt^2 - 2t dy/dt + 2y = 0, given that y1(t) = t is one solution.



My attempt:
divided equation by 1+t^2
d^2y/dt^2 - 2t/1+t^2 dy/dt + 2/1+t^2 y = 0

using u(t) eq'n given:
u(t) = exp(-integ(- 2t/t+t^2 dt)) / y1^2(t)
let x = 1+t^2
dx/dt = 2t
dt = dx/2t
u(t) = exp(integ(2t/x dx/2t)) / t^2
u(t) = exp(integ(1/x dx)) / t^2
u(t) = exp(ln x) / t^2
u(t) = e^ln(1+t^2) / t^2
u(t) = e^ln(1+t^2) / t^2
u(t) = (1+t^2) / t^2
u(t) = (1+t^2)/(t^2)

using y2(t) eq'n given:
y2(t) = t integ (u(t)dt)
y2(t) = t integ ((1+t^2)/(t^2)dt)
y2(t) = t [ integ(1/t^2 dt) + integ (1 dt)]
y2(t) = t[-1/t + t]
y2(t) = -1 + t^2

y(t) = c1y1 + c2y2
y(t) = c1t + c2[-1 + t^2]
y(t) = c1t - c2[t^2 - 1]

can someone please confirm whether i am doing this correct?

 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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sorry the title should read 2nd order homogenous eq'n, not nonhomogenous

Find the general solution of the equation:
(1+t^2)d^2y/dt^2 - 2t dy/dt + 2y = 0, given that y1(t) = t is one solution.



My attempt:
divided equation by 1+t^2
d^2y/dt^2 - 2t/1+t^2 dy/dt + 2/1+t^2 y = 0

using u(t) eq'n given:
u(t) = exp(-integ(- 2t/t+t^2 dt)) / y1^2(t)

What "u(t) eq'n" are you talking about? And why write "y1^2(t)" when you know y1= t?

let x = 1+t^2
dx/dt = 2t
dt = dx/2t
u(t) = exp(integ(2t/x dx/2t)) / t^2
u(t) = exp(integ(1/x dx)) / t^2
u(t) = exp(ln x) / t^2
u(t) = e^ln(1+t^2) / t^2
u(t) = e^ln(1+t^2) / t^2
u(t) = (1+t^2) / t^2
u(t) = (1+t^2)/(t^2)

using y2(t) eq'n given:
and what "y2(t) eq'n" were you given?
Are you using formulas in your textbook? We can't very well tell whether what you are doing is correct or not if we don't know what formulas you are using.

y2(t) = t integ (u(t)dt)
y2(t) = t integ ((1+t^2)/(t^2)dt)
y2(t) = t [ integ(1/t^2 dt) + integ (1 dt)]
y2(t) = t[-1/t + t]
y2(t) = -1 + t^2

y(t) = c1y1 + c2y2
y(t) = c1t + c2[-1 + t^2]
y(t) = c1t - c2[t^2 - 1]

can someone please confirm whether i am doing this correct?
Since you did not explain what you are doing I can't tell whether it is or is not correct. I can say, since it is obvious, that your final y satisfies the differential equation. Surely you had already seen that.
 
  • #3
162
0
sorry
the equations i used are the following:
u(t) = exp(-integ(b/a dt))
and
y2(t) = y1(t) integ (u(t) dt)
 

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