Solve for y(x) using the Fundamental Theorem of Calculus

In summary: I thought the generality of C was in the fact that it can take on any constant value. I'll definitely keep that in mind for future problems.
  • #1
Kaylee
10
0

Homework Statement


Solve the integral equation for y(x):
y(x) = 1 + ∫ { [y(t)]^2 / (1 + t^2) } dt
(integral from 0 to x)

See attached image for the equation in a nicer format.

Homework Equations


Fundamental Theorem of Calculus

The Attempt at a Solution


dy/dx = y(x)^2 / (1 + x^2)
∫ dy/y^2 = ∫ dx/(1 + x^2)
-1/y = arctan(x) + C
y = -1/arctan(x) + C

y(0) = 1 = -1/arctan(0) + C
C = 1 + 1/arctan(0)
Since arctan(0) = 0, the fraction is indeterminate.

In lectures we only did one practice example, so if anyone could give me a hand and point out where I am going wrong, it'll be greatly appreciated.
 

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  • #2
Kaylee said:

Homework Statement


Solve the integral equation for y(x):
y(x) = 1 + ∫ { [y(t)]^2 / (1 + t^2) } dt
(integral from 0 to x)

See attached image for the equation in a nicer format.

Homework Equations


Fundamental Theorem of Calculus

The Attempt at a Solution


dy/dx = y(x)^2 / (1 + x^2)
∫ dy/y^2 = ∫ dx/(1 + x^2)
-1/y = arctan(x) + C
y = -1/arctan(x) + C

y(0) = 1 = -1/arctan(0) + C
C = 1 + 1/arctan(0)
Since arctan(0) = 0, the fraction is indeterminate.

In lectures we only did one practice example, so if anyone could give me a hand and point out where I am going wrong, it'll be greatly appreciated.

Use parentheses! -1/arctan(x)+C is not the same thing as -1/(arctan(x)+C)! One is indeterminant, one is -1/C. Which one do you want?
 
  • #3
Kaylee said:

Homework Statement


Solve the integral equation for y(x):
y(x) = 1 + ∫ { [y(t)]^2 / (1 + t^2) } dt
(integral from 0 to x)

See attached image for the equation in a nicer format.

Homework Equations


Fundamental Theorem of Calculus

The Attempt at a Solution


dy/dx = y(x)^2 / (1 + x^2)
∫ dy/y^2 = ∫ dx/(1 + x^2)
-1/y = arctan(x) + C
y = -1/(arctan(x) + C)

y(0) = 1 = -1/arctan(0) + C
C = 1 + 1/arctan(0)
Since arctan(0) = 0, the fraction is indeterminate.
This is a good example of why you need to not be sloppy about using parentheses.

[Edit]: Dick wins again.
 
  • #4
Dick said:
Use parentheses! -1/arctan(x)+C is not the same thing as -1/(arctan(x)+C)! One is indeterminant, one is -1/C. Which one do you want?
I definitely want -1/C. Thank you for helping me figure it out. I knew it was something small but I couldn't see it.
 
  • #5
LCKurtz said:
This is a good example of why you need to not be sloppy about using parentheses.

[Edit]: Dick wins again.
It wasn't that I was sloppy. I commonly leave the C on the outer edge of my equations because C cannot loss its generality. But it definitely reminded me to use the generality to my advantage. Thank you to Dick and yourself for helping students with their problems.
 
  • #6
Kaylee said:
It wasn't that I was sloppy. I commonly leave the C on the outer edge of my equations because C cannot loss its generality. But it definitely reminded me to use the generality to my advantage. Thank you to Dick and yourself for helping students with their problems.

Once you've put in the C after you integrate, you have to leave it where it is. You can't just move it someplace else later on without a good algebraic reason. It doesn't have that much 'generality'. -1/(arctan(x)+C) is a general solution to your ode. -1/arctan(x)+C is NOT.
 
  • #7
Dick said:
Once you've put in the C after you integrate, you have to leave it where it is. You can't just move it someplace else later on without a good algebraic reason. It doesn't have that much 'generality'. -1/(arctan(x)+C) is a general solution to your ode. -1/arctan(x)+C is NOT.
Thanks. I didn't know that.
 

1. What is the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus is a fundamental concept in calculus that relates the concepts of differentiation and integration. It states that the integral of a function f(x) from a to b can be evaluated by finding the antiderivative of f(x) and evaluating it at the two points a and b, and then taking the difference between the two values.

2. How do you use the Fundamental Theorem of Calculus to solve for y(x)?

To solve for y(x) using the Fundamental Theorem of Calculus, you need to first find the antiderivative of the function in question. This can be done by using integration techniques such as substitution or integration by parts. Once you have the antiderivative, you can evaluate it at the two points a and b and then take the difference between the two values to find the value of y(x).

3. What are the two parts of the Fundamental Theorem of Calculus?

The first part of the Fundamental Theorem of Calculus states that if a function f(x) is continuous on the interval [a, b], then the integral of f(x) from a to b can be evaluated by finding the antiderivative of f(x) and evaluating it at the two points a and b. The second part states that if f(x) is continuous on the interval [a, b] and F(x) is any antiderivative of f(x), then the integral of f(x) from a to b is equal to F(b) - F(a).

4. Can the Fundamental Theorem of Calculus be used to evaluate any integral?

No, the Fundamental Theorem of Calculus can only be used to evaluate definite integrals, which have specific lower and upper limits of integration. It cannot be used for indefinite integrals or integrals with variable limits of integration.

5. How does the Fundamental Theorem of Calculus relate to the concept of area under a curve?

The Fundamental Theorem of Calculus can be used to find the area under a curve by using integration to find the antiderivative of the function and evaluating it at the two points a and b. This gives the value of the definite integral, which represents the area under the curve between the two points. So, the Fundamental Theorem of Calculus provides a way to calculate the area under a curve by using the concept of integration.

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