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Differential Eq. (Substitution)

  • #1
3,003
2

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!
 
Last edited:

Answers and Replies

  • #3
1,631
4

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], lets 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 lets 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?
 

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