• Support PF! Buy your school textbooks, materials and every day products Here!

Differential equation substitution method

  • Thread starter tony873004
  • Start date
  • #1
tony873004
Science Advisor
Gold Member
1,751
141
[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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,482
3
Take the derivative of y = vx with respect to x, then equate it, and you are done.
 
  • #3
gabbagabbahey
Homework Helper
Gold Member
5,002
6
[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].
 

Related Threads for: Differential equation substitution method

Replies
5
Views
3K
Replies
6
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
717
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
717
Replies
4
Views
9K
Top