Differential equation substitution method

We get \frac{dy}{dx}=v+x\frac{dv}{dx}. Substituting v=y/x, we get \frac{dy}{dx}=y/x+x\frac{d(y/x)}{dx}. Simplifying, we get \frac{dy}{dx}=\frac{y}{x}+x\frac{d(y/x)}{dx}. Since \frac{d(y/x)}{dx}=0, the final line becomes \frac{dy}{dx}=x\frac{dv}{dx}+v.
  • #1
tony873004
Science Advisor
Gold Member
1,752
143
[tex]\begin{array}{l}
\frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\
\\
\frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\
\\
{\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\
\\
\frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\
\\
\frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{x^2 }} + v^2 \\
\end{array}[/tex]

But the class notes say the final line should be
[tex]\frac{dy}{dx}=x\frac{dv}{dx}+v[/tex]
How did he get that?
 
Physics news on Phys.org
  • #2
Take the derivative of y = vx with respect to x, then equate it, and you are done.
 
  • #3
tony873004 said:
[tex]\begin{array}{l}
\frac{{dy}}{{dx}} = \frac{{4x^2 + 5xy + y^2 }}{{x^2 }} \\
\\
\frac{{dy}}{{dx}} = 4 + \frac{{5y}}{x} + \left( {\frac{y}{x}} \right)^2 \\
\\
{\rm{Let }}v = y/x\,\,\,\, \Rightarrow \,\,\,\,y = vx \\
\\
\frac{{dy}}{{dx}} = 4 + \frac{{5v}}{{xx}} + \left( {\frac{{vx}}{x}} \right)^2 \\
\\
\end{array}[/tex]

This should be [itex]\frac{dy}{dx}= 4 + \frac{{5vx}}{{x}} + \left( {\frac{{vx}}{x}} \right)^2=4+5v+v^2[/itex]

But the class notes say the final line should be
[tex]\frac{dy}{dx}=x\frac{dv}{dx}+v[/tex]
How did he get that?

Use the product rule to differentiate [itex]y=vx[/itex].
 

1. What is the substitution method in differential equations?

The substitution method in differential equations is a technique used to solve first-order differential equations by substituting a new variable or expression for the existing variable. This method is especially useful when the differential equation is not in a form that can be easily solved by other methods, such as separation of variables or using an integrating factor.

2. How does the substitution method work in solving differential equations?

In the substitution method, a new variable or expression is substituted for the existing variable in the differential equation. This new variable is chosen in such a way that it simplifies the equation and makes it easier to solve. The differential equation is then solved for the new variable, and the solution is converted back to the original variable to obtain the final solution.

3. What are the advantages of using the substitution method?

The substitution method allows us to solve differential equations that are not easily solvable by other methods. It also simplifies the equation and makes it easier to solve, especially when the equation involves complex functions or expressions. Additionally, it can be used to solve a wide range of differential equations, including non-linear and non-separable equations.

4. Are there any limitations to using the substitution method?

One limitation of the substitution method is that it may not always be possible to find a suitable substitution that simplifies the equation. In some cases, the substitution may lead to a more complicated equation, making it difficult to solve. Additionally, this method may not be applicable to higher-order differential equations.

5. Can the substitution method be used for both homogeneous and non-homogeneous differential equations?

Yes, the substitution method can be used to solve both homogeneous and non-homogeneous differential equations. However, the choice of substitution may differ depending on the type of equation. For homogeneous equations, a substitution involving the ratio of the dependent and independent variables is often used, while for non-homogeneous equations, a substitution involving the dependent variable itself is more common.

Similar threads

  • Calculus and Beyond Homework Help
Replies
19
Views
766
  • Calculus and Beyond Homework Help
Replies
6
Views
748
  • Calculus and Beyond Homework Help
Replies
21
Views
828
  • Calculus and Beyond Homework Help
Replies
10
Views
422
  • Calculus and Beyond Homework Help
Replies
1
Views
820
  • Calculus and Beyond Homework Help
Replies
25
Views
320
  • Calculus and Beyond Homework Help
Replies
6
Views
842
  • Calculus and Beyond Homework Help
Replies
20
Views
448
  • Calculus and Beyond Homework Help
Replies
2
Views
721
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
Back
Top