Solving a Quadratic Equation with Initial Conditions

In summary, the question is about finding the initial conditions for a separable ODE with a quadratic solution, and the nature of the solutions can be determined by the discriminant.
  • #1
jacko200
2
0
Ok, the question I have is attached. What I have done is found the general solution:

(Question was seperable, so it was easy)
y = x^2 - 2x + c

But I don't kwow what/how to do is finding the initial conditions where there are:
(a) No solutions
(b) more than one solution
(c) precisely one solutions

Thanks
 

Attachments

  • q8.JPG
    q8.JPG
    22.1 KB · Views: 350
Physics news on Phys.org
  • #2
I don't see your ODE, but if that quadratic is your solution, it only takes algebra to solve the problem. Recall your quadratic equation. Your discriminant (the stuff inside the square root) will tell you about the nature of the solution. Just in case you forgot.

1) Discriminant is positive: more than one solution
2) Discriminant is zero: one solution
3) Discriminant is negative: no real solution (imaginary solutions)

Your "initial condition" is the value of "c".
 

Related to Solving a Quadratic Equation with Initial Conditions

What is a quadratic equation?

A quadratic equation is a mathematical expression in the form of ax^2 + bx + c = 0, where x is the variable and a, b, and c are constants. It is a second-degree polynomial equation, meaning the highest exponent of the variable is 2.

How do I solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula, factoring, or completing the square methods. The quadratic formula is the most commonly used method and is given by x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients in the equation.

What are initial conditions in a quadratic equation?

Initial conditions refer to the given values or constraints in a problem that involve a quadratic equation. These can include the initial position, velocity, or time in a physics problem, or the initial investment, interest rate, or time period in a financial problem.

Why do we need to consider initial conditions when solving a quadratic equation?

Initial conditions are important because they provide specific information about the problem and help us find the unique solution to the quadratic equation. Without these conditions, there may be multiple solutions or no solution at all.

What are some real-life applications of solving a quadratic equation with initial conditions?

Quadratic equations with initial conditions are commonly used in physics, engineering, and finance. For example, they can be used to model the motion of a projectile, calculate the optimal production level for a company, or determine the time it takes for an investment to reach a certain value.

Similar threads

Replies
4
Views
600
  • Calculus and Beyond Homework Help
Replies
8
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
Replies
12
Views
477
  • Calculus and Beyond Homework Help
Replies
1
Views
318
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
636
  • Calculus and Beyond Homework Help
Replies
8
Views
814
Back
Top