Differential Eq. (Substitution)

In summary, thehomework statement has the equation xdx+(y-2x)dy=0. The equation can be solved by either y=ux or x=vy. However, the latter seemed easier. The equation can be solved by taking the subtrahends v and y and combining them to get v+y\frac{dv}{dy}+\frac{1}{v}. This is separable and the student should be fine from here on.
  • #1
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Homework Statement


Solve by making an appropriate substitution. I am given the homogeneous DE:[tex]xdx+(y-2x)dy=0[/tex]

Now we have bee using either y=ux or x=vy. . . I tried both, but the latter seemed easier.

[tex]x\frac{dx}{dy}+y-2x=0[/tex] letting x=vy and dx/dy=v+y*dy/dv

[tex]vy(v+y\frac{dy}{dv})+y-2vy=0[/tex]

[tex]v^2+y^2\frac{dy}{dv}+y-2vy=0[/tex]

Here is where I get stumped. . . this is supposed to be separable now right? Because I can't seem to see it.

A hint would be swell!
 
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  • #2
Saladsamurai said:
dx/dy=v+y*dy/dv
I think you need to check that again.
 
  • #3
Saladsamurai said:

Homework Statement


Solve by making an appropriate substitution. I am given the homogeneous DE:


[tex]xdx+(y-2x)dy=0[/tex]

Now we have bee using either y=ux or x=vy. . . I tried both, but the latter seemed easier.

[tex]x\frac{dx}{dy}+y-2x=0[/tex] letting x=vy and dx/dy=v+y*dy/dv

[tex]vy(v+y\frac{dy}{dv})+y-2vy=0[/tex]

[tex]v^2+y^2\frac{dy}{dv}+y-2vy=0[/tex]

Here is where I get stumped. . . this is supposed to be separable now right? Because I can't seem to see it.

A hint would be swell!

[tex]x\frac{dx}{dy}+y-2x=0[/tex], let's divide by x to get

[tex]\frac{dx}{dy}+\frac{y}{x}=2[/tex] or

[tex]\frac{dx}{dy}+(\frac{x}{y})^{-1}=2[/tex] now let's take the sub

[tex] v= \frac{x}{y}, x=vy, \frac{dx}{dy}=v+y\frac{dv}{dy}[/tex],(note: you made a mistake here) now let's go back and substitute we get

[tex]v+y\frac{dv}{dy}+\frac{1}{v}=2[/tex] so we get

[tex]y\frac{dv}{dy}+\frac{v^{2}+1}{v}=2[/tex]

[tex]y\frac{dv}{dy}=\frac{2v-v^{2}-1}{v}=-\frac{(v-1)^{2}}{v}[/tex]

Now this is separable and i think you will be fine from here on, right?
 

1. What is substitution in differential equations?

Substitution in differential equations involves replacing one or more variables with a new variable or expression in order to simplify the equation or make it solvable.

2. When should substitution be used in solving differential equations?

Substitution should be used when the equation cannot be solved using traditional methods, such as separation of variables or integrating factors.

3. How do you perform substitution in differential equations?

To perform substitution, you must identify which variable(s) can be replaced and with what expression. Then, make the appropriate substitution and solve the resulting equation.

4. What are the benefits of using substitution in solving differential equations?

Substitution can help make a difficult or unsolvable equation more manageable and solvable. It can also reveal patterns or relationships that were not initially apparent in the original equation.

5. Are there any limitations to using substitution in differential equations?

Yes, substitution may not always be successful in solving an equation, especially if the new variable or expression chosen is not appropriate for the equation. It also may not work for more complex or non-linear equations.

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