Differential equations - 2nd order nonhomogenous eq'n

In summary, the given differential equation (1+t^2)d^2y/dt^2 - 2t dy/dt + 2y = 0 can be solved by dividing by 1+t^2 and using the equations u(t) = exp(-integ(b/a dt)) and y2(t) = y1(t) integ (u(t) dt) - integ(y1(t) u(t) dt). The general solution is y(t) = c1t - c2[t^2 - 1], where c1 and c2 are arbitrary constants.
  • #1
braindead101
162
0
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:
Physics news on Phys.org
  • #2
braindead101 said:
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
sorry
the equations i used are the following:
u(t) = exp(-integ(b/a dt))
and
y2(t) = y1(t) integ (u(t) dt)
 

1. What is a 2nd order nonhomogenous differential equation?

A 2nd order nonhomogenous differential equation is a mathematical equation that involves derivatives of a dependent variable with respect to an independent variable, where the equation is not equal to zero and cannot be simplified into a linear form. It is considered nonhomogenous because it includes a function that is not equal to zero.

2. How is a 2nd order nonhomogenous differential equation solved?

A 2nd order nonhomogenous differential equation is typically solved using a combination of techniques, such as the method of undetermined coefficients, variation of parameters, or the Laplace transform. These methods involve finding a particular solution and a complementary solution, which are then added together to form the general solution.

3. What is the difference between a 2nd order nonhomogenous and homogenous differential equation?

The main difference between a 2nd order nonhomogenous and homogenous differential equation is the presence of a function that is not equal to zero in the nonhomogenous equation. In contrast, a homogenous equation has a function that is equal to zero, making it easier to solve as it can be reduced to a linear form.

4. What are some real-world applications of 2nd order nonhomogenous differential equations?

2nd order nonhomogenous differential equations have many applications in engineering, physics, and other fields of science. Some examples include modeling the motion of a spring-mass system, analyzing electrical circuits, and predicting the growth of a population over time.

5. Can 2nd order nonhomogenous differential equations have multiple solutions?

Yes, 2nd order nonhomogenous differential equations can have multiple solutions. This is because the general solution of a nonhomogenous equation includes a complementary solution and a particular solution, which can take on different forms depending on the specific equation and initial conditions. However, the form of the general solution is unique and can be written in terms of arbitrary constants.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
544
  • Calculus and Beyond Homework Help
Replies
8
Views
189
  • Calculus and Beyond Homework Help
Replies
2
Views
490
  • Calculus and Beyond Homework Help
Replies
12
Views
972
  • Calculus and Beyond Homework Help
Replies
0
Views
122
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
686
  • Calculus and Beyond Homework Help
Replies
7
Views
237
  • Calculus and Beyond Homework Help
Replies
5
Views
884
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
Back
Top