Determine the general solution of QL PDE

[bugatti79]

Homework Statement

1) Determine the general solution of the equation

2) Use implict differentiation to verify that your solution satisfies the given PDE

Homework Equations

u u_x-y u_y=y

The Attempt at a Solution

\frac{dx}{u}=\frac{dy}{-y}=\frac{du}{y}

Take the second two

\int-dy=\int du \implies u=-y+A

Taking the first two

\frac{dx}{(-y+A)}=\frac{dy}{-y} \implies dx=\frac{(-y+A)dy}{-y}

Integrating gives

x=y-A \ln(y) + f(A) but f(A)=u+y therefore the general solution implicitly is

x=y-A \ln(y) + u+y


1) How am I doing?
2) I don't know how to do second question assuming above is correct
3) How do I create the tags automatically?

Thanks

 

[bugatti79]

Folks,

I have also posted this query at

http://www.mathhelpforum.com/math-help/f59/determine-general-solution-ql-pde-185703.html"

I will keep each forum informed. No luck yet.

Thanks

 

[bugatti79]

Homework Statement

1) Determine the general solution of the equation

2) Use implict differentiation to verify that your solution satisfies the given PDE

Homework Equations

u u_x-y u_y=y

The Attempt at a Solution

\frac{dx}{u}=\frac{dy}{-y}=\frac{du}{y}

Take the second two

\int-dy=\int du \implies u=-y+A

Taking the first two

\frac{dx}{(-y+A)}=\frac{dy}{-y} \implies dx=\frac{(-y+A)dy}{-y}

Integrating gives

x=y-A \ln(y) + f(A) but f(A)=u+y therefore the general solution implicitly is

x=y-A \ln(y) + f(u+y)


1) How am I doing?
2) I don't know how to do second question assuming above is correct
3) How do I create the tags automatically?

Thanks

I realized I left out the function f symbol as highlighted above. Any ideas?

 

[bugatti79]

This is solved...See link in post 2

Cheers