Differential Equations: If y(1) =3 then y(1/2) = ?

Then, use a substitution u = ln(t) to reduce it to a separable D.E. Finally, solve for y(t) and plug in the initial condition y(1) = 3 to find the value of y(1/2). In summary, to find y(1/2), factor out t and use a substitution u = ln(t) to reduce the D.E. to a separable form, solve for y(t), and plug in the initial condition y(1) = 3.
  • #1
Northbysouth
249
2

Homework Statement


If y satisfies the differential equation:

ty'(t) + tln(t)y(t) = 0

and y(1) = 3 then y(1/2) = ?

I have attached an image of the question with the possible solutions:


Homework Equations





The Attempt at a Solution



Initially I tried plugging in t=1 and y(1) = 3 giving me:

y'(1)+3ln(1) = 0

Hence y'(1) = 0

I did the same for t=1/2

1/2[y'(1/2) + ln(1/2)y(1/2)] = 0

But this doesn't seem to help me. Any suggestions would be appreciated.
 

Attachments

  • math 2214 quiz 4 2.png
    math 2214 quiz 4 2.png
    7.5 KB · Views: 477
Physics news on Phys.org
  • #2
Why don't you just solve the DE and put in the boundary conditions? It looks to be separable and linear so you have a couple of options.
 
  • #3
Northbysouth said:

Homework Statement


If y satisfies the differential equation:

ty'(t) + tln(t)y(t) = 0

and y(1) = 3 then y(1/2) = ?

I have attached an image of the question with the possible solutions:

Homework Equations



The Attempt at a Solution



Initially I tried plugging in t=1 and y(1) = 3 giving me:

y'(1)+3ln(1) = 0

Hence y'(1) = 0

I did the same for t=1/2

1/2[y'(1/2) + ln(1/2)y(1/2)] = 0

But this doesn't seem to help me. Any suggestions would be appreciated.
To help you solve the D.E., factor out t .
 

1. What is the method for solving a differential equation?

There are several methods for solving a differential equation, including separation of variables, integrating factors, and the method of undetermined coefficients. The specific method used depends on the type of differential equation and its initial conditions.

2. How do you find the general solution of a differential equation?

The general solution of a differential equation is the set of all possible solutions. To find the general solution, the differential equation must be solved using one of the methods mentioned above. The resulting equation will contain a constant, which can take on any value, thus representing the general solution.

3. What are initial conditions in a differential equation?

Initial conditions are specific values or conditions given for the dependent variable and its derivatives at a specific point. These conditions are used to determine the particular solution of the differential equation.

4. How do you use initial conditions to find the particular solution of a differential equation?

To find the particular solution of a differential equation, plug the initial conditions into the general solution and solve for the constant. This will give a specific solution that satisfies the given initial conditions.

5. Can a differential equation have more than one solution?

Yes, a differential equation can have an infinite number of solutions. However, for a specific set of initial conditions, there will only be one particular solution that satisfies those conditions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
191
  • Calculus and Beyond Homework Help
Replies
2
Views
733
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
688
  • Calculus and Beyond Homework Help
Replies
6
Views
760
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
324
  • Calculus and Beyond Homework Help
Replies
7
Views
284
  • Calculus and Beyond Homework Help
Replies
5
Views
289
  • Calculus and Beyond Homework Help
Replies
3
Views
571
Back
Top