Efficient Techniques for Solving Second Order Differential Equations

In summary, the conversation discusses finding the solution for part (b) of a problem involving characteristic equations and ODEs. The participants discuss whether to find the characteristic equation and then the solution y2, or to just integrate the expression from part (a). They also discuss how to derive y2 and what to do with the resulting derivatives. The conversation ends with the suggestion to use the fact that y1 is a solution of the ODE to solve for the unknowns p(x) and q(x).
  • #1
nysnacc
184
3

Homework Statement


upload_2016-10-17_22-21-28.png


Homework Equations

The Attempt at a Solution


Should I find the characteristic equation then find the solution y2? Or just integrate the expression of part (a)??

Part B, just find the derivatives and plug into the ode
 
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  • #2
nysnacc said:
Should I find the characteristic equation then find the solution y2? Or just integrate the expression of part (a)??
yes you can put the expression in the ODE and use the fact that ##y_{1}## is a particular solution ...
 
  • #3
So, I calculate the y2, y2' and y2" in terms of y1, and plug into ODE (1st line)??
 
  • #4
yes, as hint I show you how to derive ##y_{2}(x)## respect ##x##. We observe that we have a product so:

##y_{2}'(x)=y_{1}'(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx + y_{1}(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}##

try for ##y_{2}''## ...
 
  • #5
d/dx ##y_{1}(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}##

= -1/y1^2 ##e^{-\int p(x)dx}## +1/y p(x) ##e^{-\int p(x)dx}##
 
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  • #6
After hving the y, y', y", what should I do?
 
  • #7
nysnacc said:
d/dx y1(x)e−∫p(x)dxy21(x)y_{1}(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}

= -1/y1^2 e^{-\int p(x)dx}}e^{-\int p(x)dx}} +1/y p(x) e^{-\int p(x)dx}}e^{-\int p(x)dx}}
wait this not ##y_{2}''(x)## this is only a part of the second derivative, you forgot to derive the term ##y_{1}'(x)\int\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx## ...
 
  • #8
Yes, you're right, but is this part correct tho?
 
  • #9
yes but there is ##-## instead ##+## in the middle...
 
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  • #10
oh right, because of ∫ -p(x) :)
 
  • #11
Then I plug into the ODE?? but I have 2 unknowns p(x) and q(x) ... what should I do?
 
  • #12
If your calculation are correct you must have this:

##y_{1}''(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx +y_{1}'(x)\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}+\frac{-p(x)y(x)e^{-\int p(x)dx}-y_{1}'(x)e^{-\int p(x)dx}}{y_{1}^{2}(x)}+##
##+p(x)y_{1}'\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx+p(x)\frac{e^{-\int p(x)dx}}{y_{1}(x)}+q(x)y_{1}(x)\int \frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx=0##
 
  • #13
that is ##\left[y_{1}''(x)+p(x)y_{1}'(x)+q(x)y_{1}(x)\right]\int\frac{e^{-\int p(x)dx}}{y_{1}^{2}(x)}dx=0##, you can simplify and this prove the first part ...
 
  • #14
nysnacc said:
Then I plug into the ODE?? but I have 2 unknowns p(x) and q(x) ... what should I do?
You need to use the fact that ##y_1## is a solution of the ODE.
 
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  • #15
thank you
 

What is the method of reduction of order?

The method of reduction of order is a technique used in mathematics and physics to simplify and solve differential equations. It involves reducing a higher-order differential equation to a lower-order equation by making a substitution, which allows for an easier solution to be found.

When is the method of reduction of order used?

The method of reduction of order is typically used when solving second-order linear differential equations with constant coefficients. It can also be used for higher-order equations, but the process can become more complicated.

How does the method of reduction of order work?

The method of reduction of order involves making a substitution of the form y = ux, where u is a new function of x. This reduces the order of the differential equation by one and transforms it into an equation with separable variables, which can be solved using standard techniques.

What are the steps involved in the method of reduction of order?

The steps for the method of reduction of order are as follows: 1. Assume a solution of the form y = ux, where u is a new function of x. 2. Substitute this solution into the original differential equation. 3. Simplify the resulting equation and solve for u. 4. Use the solution for u to find the solution for y.

What are the advantages of using the method of reduction of order?

The method of reduction of order allows for solving higher-order differential equations using standard techniques for first-order equations. It also simplifies the process of finding a particular solution and can be applied to a wide range of differential equations in physics and engineering.

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