How do I start this ODE substitution problem?

In summary, using the substitution v=x+y, the equation (x+y)y' = x-y can be rewritten as x^2 - 2xy - y^2 = C, as shown in the back of the book. Through a series of substitutions and integrations, the equation can be simplified to x^2 - 2xy - y^2 = C, matching the solution in the back of the book.
  • #1
tony873004
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Homework Statement


[tex]\left( {x + y} \right)y' = x - y[/tex]


Homework Equations


Back of book:
[tex]x^2 - 2xy - y^2 = C[/tex]


The Attempt at a Solution


I'm not sure how to start this problem. In the examples in the book, they make a substitution, v=something, and all that was left were v's and x's, or v's and y's. Usually it was an expression that appeared twice, so making a substution made sense. But if I set v=x+y, or v=x-y, I'm still left with v's, x's & y's. How do I start this one?

Thanks in advance!
 
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  • #2
You are giving up too fast. v=x+y will work. Just try harder. Notice e.g. y=v-x, so x-y=2x-v.
 
  • #3
Thanks, I think I'm on a better path now. I get this far:
[tex]\begin{array}{l}
\left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y\,\,\,\, \Rightarrow \,\,\,\,y = v - x \\
\\
vy' = x - y\,\,\,\, \Rightarrow \,\,\,\,v\frac{{dy}}{{dx}} = x - v - x\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} = - v \\
\end{array}[/tex]

This is where the example in the book just skips a step, so I don't know what to do next.

edit*** don't respond yet, let me try what u suggested
 
  • #4
Be careful. On the second line you should have x-(v-x). That doesn't simplify to -v. Just a warning.
 
  • #5
It's starting to look like the back of the book! Am I making this harder than it needs to be?
[tex]
\begin{array}{l}
\left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y\,\,\,\, \Rightarrow \,\,\,\,y = v - x\,\,\,\, \Rightarrow \,\,\,\,x - y = 2x - v \\
\\
vy' = 2x - v \\
\\
vy' + v = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {y' + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {y' + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{d\left( {v - x} \right)}}{{dx}} + 1} \right) \\
\\
x = v\left( {\frac{{dv - dx}}{{dx}} + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{dv}}{{dx}} - \frac{{dx}}{{dx}} + 1} \right)\,\,\,\, \Rightarrow \,\,\,\,x = v\left( {\frac{{dv}}{{dx}} - 1 + 1} \right)\,\,\,\, \Rightarrow \,\,\,\, \\
\\
x = v\left( {\frac{{dv}}{{dx}}} \right)\,\,\,\, \Rightarrow \,\,\,\,vdv = xdx \\
\\
\int_{}^{} {v\,dv} = \int_{}^{} {x\,dx} \\
\\
\frac{{v^2 }}{2} + C_1 = \frac{{x^2 }}{2} + C_2 \\
\\
\frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C_2 - C_1 ,\,\,\,\,\,\,\,C = \,C_2 - C_1 \\
\\
\frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} = \frac{{x^2 }}{2} + \frac{C}{2}\,\,\,\, \Rightarrow \,\,\,\,x^2 + 2xy + y^2 = x^2 + C\,\,\,\, \Rightarrow \,\, \\
\\
\frac{{x^2 + 2xy + y^2 }}{{x^2 }} = x^2 + C\,\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{2xy + y^2 }}{{x^2 }} + 1 = x^2 + C \\
\end{array}
[/tex]

[tex]
\begin{array}{l}
\frac{{\left( {x + y} \right)^2 }}{2} = \frac{{x^2 }}{2} + C\,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} = \frac{{x^2 }}{2} + \frac{C}{2}\,\,\,\, \Rightarrow \,\,\,\,x^2 + 2xy + y^2 = x^2 + C\,\,\,\, \Rightarrow \,\, \\
\\
\frac{{x^2 + 2xy + y^2 }}{{x^2 }} = x^2 + C\,\,\,\,\,\, \Rightarrow \,\,\,\,\frac{{2xy + y^2 }}{{x^2 }} + 1 = x^2 + C\,\, \\
\end{array}
[/tex]
 
  • #6
Dick said:
Be careful. On the second line you should have x-(v-x). That doesn't simplify to -v. Just a warning.

That might be where my mistake lies. Gotta run to class now. I'll try that in a few hours. Thanks!
 
  • #7
tony873004 said:
That might be where my mistake lies. Gotta run to class now. I'll try that in a few hours. Thanks!

Wow. So many steps! But you dropped a 2 on the third line.
 
  • #8
Dick said:
Wow. So many steps! But you dropped a 2 on the third line.
That 2 turned out to be important :)

[tex]
\begin{array}{l}
\left( {x + y} \right)y' = x - y,\,\,\,\,\,\,\,\,\,v = x + y,\,\,\,\,\,\,\,y = v - x,\,\,\,\,\,\,\,\,x - y = x - \left( {v - x} \right) = \,\,\,\, \Rightarrow \,\,\,\,x - y = x - v + x\,\,\,\, \Rightarrow \,\,\,\,x - y = 2x - v \\
\\
v\left( {v - x} \right)^\prime = 2x - y\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} = 2x - v\,\,\,\, \Rightarrow \,\,\,\,v\frac{{d\left( {v - x} \right)}}{{dx}} + v = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{d\left( {v - x} \right)}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v\left( {\frac{{dv - dx}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{dv}}{{dx}} - \frac{{dx}}{{dx}} + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\left( {\frac{{dv}}{{dx}} - 1 + 1} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v\left( {\frac{{dv}}{{dx}}} \right) = 2x\,\,\,\, \Rightarrow \,\,\,\,v\,\,dv = 2x\,\,dx \\
\\
\int_{}^{} {v\,\,dv} = \int_{}^{} {2x\,\,dx} \\
\\
\frac{{v^2 }}{2} + C_1 = \frac{{2x^2 }}{2} + C_2 \\
\end{array}
\[/tex]

[tex]
\begin{array}{l}
\frac{{v^2 }}{2} = x^2 + C_2 - C_1 ,\,\,\,\,\,\,\,C_3 = C_2 - C_1 \\
\\
\frac{{v^2 }}{2} = x^2 + C_3 \,\,\,\, \Rightarrow \,\,\,\,\frac{{\left( {x + y} \right)^2 }}{2} - x^2 = C_3 \,\,\,\, \Rightarrow \,\,\,\,\frac{{x^2 + 2xy + y^2 }}{2} - x^2 = C_3 \,\,\,\, \Rightarrow \,\,\,\, \\
\\
x^2 + 2xy + y^2 - 2x^2 = 2C_3 \\
\\
- x^2 + 2xy + y^2 = 2C_3 ,\,\,\,\,\,\,\,\,\,\,\,\,\, - C = 2C_3 \, \\
\\
- x^2 + 2xy + y^2 = - C \\
\\
x^2 - 2xy - y^2 = C \\
\end{array}
\[/tex]


Same as the back of the book. Thanks!
 

1. What is the Ode substitution method?

The Ode substitution method is a mathematical technique used to solve ordinary differential equations (ODEs). It involves substituting a new variable for the independent variable in the equation, which can help simplify and solve the ODE.

2. When is the Ode substitution method used?

The Ode substitution method is used when an ODE cannot be solved using other methods such as separation of variables or variation of parameters. It is also useful when the ODE is non-linear or contains complicated functions.

3. How does the Ode substitution method work?

The Ode substitution method involves substituting a new variable into the ODE, which transforms the equation into a new form. This new form can then be solved using integration and algebraic techniques.

4. What is the advantage of using the Ode substitution method?

The Ode substitution method can sometimes make it easier to solve ODEs, especially when the original equation is complex or non-linear. It can also help to reveal patterns and relationships within the equation.

5. Are there any limitations to the Ode substitution method?

The Ode substitution method may not always work for all types of ODEs. It may also lead to more complicated solutions, and it can be time-consuming. It is important to check the solution obtained using the Ode substitution method to ensure its validity.

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