How to integrate this function?

In summary, the power rule can be used to integrate functions with variables raised to a power by adding 1 to the power and dividing by the new power. Substitution can also be used to integrate functions with nested functions. Trigonometric functions can be integrated using identities and basic rules, potentially requiring substitution or integration by parts. Definite integration finds the numerical value over a specific interval, while indefinite integration finds the general antiderivative. Software can also be used for integration, but understanding the concepts is important for effective use.
  • #1
vanceEE
109
2

Homework Statement


$$xy'' = y' + (y')^{3}$$

Homework Equations


$$ y'' = p' = \frac{dp}{dx}$$
$$ y' = p$$

The Attempt at a Solution


$$xy'' = y' + (y')^{3}$$
$$ x*\frac{dp}{dx} = p + p^{3} $$
$$ \frac{dx}{x} = \frac{dp}{p+p^3} $$
$$ ln x + C = ∫\frac{dp}{p(1+p^2)} $$
$$ u = p^{2} $$ $$du = 2p dp$$
$$ ln x + C = ∫\frac{1}{p(1+p^2)}*\frac{du}{2p} $$
$$ ln x + C = ∫\frac{du}{2p^2(1+p^2)} $$
$$ ln x + C = ∫\frac{du}{2u(1+u)} $$
$$ln x + C = \frac{1}{2}∫\frac{du}{u} - \frac{1}{2}∫\frac{du}{1+u} $$
$$ln x + C = \frac{1}{2}ln (\frac{p^2}{p^2+1}) $$
$$ Dx^2 = \frac{p^2}{p^2+1} $$
$$ p^2 = \frac{Dx^2}{1-Dx^2} $$
$$ y' = \frac{√(Dx^2)}{√(1-Dx^2)} $$

How do I integrate $$ y' = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
 
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  • #2
vanceEE said:

Homework Statement


$$xy'' = y' + (y')^{3}$$


Homework Equations


$$ y'' = p' = \frac{dp}{dx}$$
$$ y' = p$$



The Attempt at a Solution


$$xy'' = y' + (y')^{3}$$
$$ x*\frac{dp}{dx} = p + p^{3} $$
$$ \frac{dx}{x} = \frac{dp}{p+p^3} $$
$$ ln x + C = ∫\frac{dp}{p(1+p^2)} $$
$$ u = p^{2} $$ $$du = 2p dp$$

You don't want a u substitution here. Try partial fractions$$
\frac 1 {p(1+p^2)}=\frac A p + \frac {Bp+C}{1+p^2}$$
 
  • #3
LCKurtz said:
You don't want a u substitution here. Try partial fractions$$
\frac 1 {p(1+p^2)}=\frac A p + \frac {Bp+C}{1+p^2}$$

I used partial fractions after u-substitution.
$$ ∫\frac{dx}{x} = ∫\frac{du}{2u(1+u)} $$
$$ \frac{A}{2u} + \frac{B}{1+u} = \frac{1}{2u(1+u)} $$
$$ A(1+u) + B(2u) = 1 $$
$$ A = 1 $$ and $$ B = -\frac{1}{2} $$
$$ ∫\frac{dx}{x} = ∫\frac{du}{2u} - \frac{1}{2}∫\frac{du}{1+u} $$
$$ ∫\frac{dx}{x} = \frac{1}{2} ∫ \frac{du}{u} - \frac{1}{2} ∫ \frac{du}{1+u} $$
$$ ln x + C = \frac{1}{2} ln(p^2) -\frac{1}{2}ln(p^2+1) $$
Therefore, $$ Dx^2 = \frac{p^2}{p^2+1} $$
Since $$ ∫\frac{du}{2u(1+u)} \equiv∫\frac{dp}{p(1+p^2)}$$
$$ du = 2p dp → u = p^2 $$
$$\frac{du}{2u(1+u)} \equiv \frac{2pdp}{2p^2(1+p^2)} \equiv \frac{dp}{p(1+p^2)} $$
 
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  • #4
Can someone show me the steps to integrating $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
Attempt:
$$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
$$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
Should I integrate by parts?
 
Last edited:
  • #5
Is this correct?
$$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
$$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
$$ y = √(D)∫\frac{x}{√(1-Dx^2)} dx $$
$$ u = 1-Dx^2 $$
$$ du = -2Dx $$
$$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
$$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
$$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
$$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
$$ y = E -\frac{√(1-Dx^2)}{√(D)} $$
 
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  • #6
vanceEE said:
Can someone show me the steps to integrating $$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} ?$$
How about x = Bsin(θ) for some suitable B?
 
  • #7
vanceEE said:
Is this correct?
$$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
$$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
$$ y = √(D)∫\frac{x}{√(1-Dx^2)} dx $$
$$ u = 1-Dx^2 $$
$$ du = -2Dx $$
The above should be du = -2Dx dx.
vanceEE said:
$$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
$$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
$$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
$$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
$$ y = E -\frac{√(1-Dx^2)}{√(D)} $$
Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.
 
  • #8
vanceEE said:
Is this correct?
$$ \frac{dy}{dx} = \frac{√(Dx^2)}{√(1-Dx^2)} $$
$$ dy = \frac{√(Dx^2)}{√(1-Dx^2)} dx$$
$$ y = ∫\frac{√(D)x}{√(1-Dx^2)} dx $$
$$ y = √(D)∫\frac{x}{√(1-Dx^2)} dx $$
$$ u = 1-Dx^2 $$
$$ du = -2Dx $$
The above should be du = -2Dx dx.
vanceEE said:
$$ y = -\frac{1}{2√(D)}∫\frac{du}{√(u)} $$
$$ y = -\frac{1}{2√(D)}* 2√(u) + E $$
$$ y = -\frac{1}{2√(D)}* 2√(1-Dx^2) + E $$
$$ y = E -\frac{2√(1-Dx^2)}{2√(D)} $$
$$ y = E -\frac{√(1-Dx^2)}{√(D)} $$
Because of the factor of x in the numerator, an ordinary substitution works here. Other than the minor error I pointed out, I don't see anything else wrong, but I didn't check that closely. At this point, you can verify that your work is correct by differentiating to verify that your solution satisifies the differential equation. You don't need us to do that for you.
 
  • #9
Mark44 said:
The above should be du = -2Dx dx

Thank you for pointing this out for me Mark44 :-)
 

1. How do I integrate a function using the power rule?

The power rule states that when integrating a function with a variable raised to a power, we add 1 to the power and divide by the new power. For example, the integral of x^2 would be (x^3)/3 + C.

2. Can I use substitution to integrate a function?

Yes, substitution can be used to integrate a function when the integrand contains a function within a function. This method involves substituting a new variable in place of the inner function and then integrating as usual.

3. How do I integrate a trigonometric function?

To integrate a trigonometric function, you can use the trigonometric identities and the basic integration rules. You may also need to use substitution or integration by parts depending on the specific function.

4. What is the difference between definite and indefinite integration?

Definite integration involves finding the exact numerical value of the integral over a specific interval, while indefinite integration involves finding the general antiderivative of a function without specifying a range.

5. Can I use software to integrate a function?

Yes, there are many software programs and online tools available that can integrate functions for you. However, it is important to understand the concepts and methods of integration in order to use these tools effectively.

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