Solving a Second Order Differential Equation: $$xy''-y'=3x^2$$

In summary, we discussed a differential equation and its various forms, including using a substitution to simplify the equation. We also looked at how to find the integrating factor and how to use the product rule to find the derivative of a product. We concluded by discussing the general form of an equation after multiplying by the integrating factor.
  • #1
vanceEE
109
2
$$ xy'' - y' = 3x^{2} $$
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
after multiplying by the integrating factor we get..
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p =3 $$
so $$ [\frac{1}{x}p]' = 3? $$

I know that these two below are equal, but can someone please show HOW
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$

Thank you!
 
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  • #2
vanceEE said:
$$ xy'' - y' = 3x^{2} $$
$$ y' = p $$
$$ y'' = p' $$
$$ xp' - p =3x^{2} $$
$$ p' - \frac{1}{x}p = 3x $$
after multiplying by the integrating factor we get..
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p =3 $$
so $$ [\frac{1}{x}p]' = 3? $$

I know that these two below are equal, but can someone please show HOW
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$

Thank you!

You want to find d/dx of the product (1/x)*p(x). Use the product rule.
 
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  • #3
vanceEE said:
I know that these two below are equal, but can someone please show HOW
$$ \frac{1}{x}p' - \frac{1}{x^{2}}p $$ equals $$[\frac{1}{x}p]' $$

Thank you!

Use the product rule on that last equation.
 
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Likes 1 person
  • #4
Awesome, I now understand that part. Now for any equation $$y' + a(x)y = b(x)$$ after multiplying the integrating factor, will my left side always be... $$ [e^{∫a(x) dx} * y]' ?$$
 
Last edited:
  • #5
I think you can answer your own question. is$$
\left(e^{∫a(x) dx} * y' + e^{∫a(x) dx} a(x)y\right)$$
the derivative of $$
e^{∫a(x) dx} * y$$
 
  • #6
yes, thank you LCKurtz!
 

1. What is a second order differential equation?

A second order differential equation is an equation that involves the second derivative of a function. It is represented in the form: $$f''(x) = g(x)$$ where f is the function and g is a known function of x.

2. How do I solve a second order differential equation?

To solve a second order differential equation, you need to find the general solution by integrating the given equation twice. Then, you can use initial conditions to find the particular solution.

3. What are initial conditions?

Initial conditions are values given to the function and its derivatives at a specific point. They are used to find the particular solution of a differential equation.

4. What is the general solution of a second order differential equation?

The general solution of a second order differential equation is the solution that includes two arbitrary constants. It is obtained by integrating the given equation twice.

5. How do I check if my solution to a second order differential equation is correct?

You can check your solution by plugging it into the original equation. If it satisfies the equation, it is the correct solution. You can also confirm by checking if it satisfies the initial conditions given in the problem.

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