How to solve $2f-f_u=g(v)$?

Question

How to solve $2f-f_u=g(v)$, where $f=f(u,v)$ ?

somehow it is similiar to a case in ODE's (integrating factor method) but here we have more than $1$ variable and RHS is a function of $v$.

for example here ''After integration we get'', how did he integrate ?

Couldn't you divide by $f_u$, add $1$ and integrate with respect to $u$? — YoTengoUnLCD, Mar 04 '16 at 16:18

Friedrich Philipp · Accepted Answer · 2016-03-04T17:18:24.313

1

Fix some $v$. Then you have $f_u = 2f - g(v)$, which is a (very simple) linear ODE in $u$ (namely $y' = 2y - g$, where $g$ is a constant). Solve it and you get $f(u,v)$ for this particular $v$. But since $v$ was arbitrary, you get $f(u,v)$. The solution is $f(u,v) = \frac 1 2g(v) + c(v)e^{2u}$, where $c$ is some function depending on $v$.

edited Mar 04 '16 at 17:18

answered Mar 04 '16 at 16:34

Friedrich Philipp

4,051

Thanks, but in the link I gave did you see there, after integrating, we have also another function $A(v)$, so the solution should be $f(u,v)=1/2g(v)+ce^{2u}+A(v)$, or not ? – ketum Mar 04 '16 at 17:06
The "constant" appearing here ($c$) should be a function of $v$, right? – YoTengoUnLCD Mar 04 '16 at 17:16
Yes, exactly that, YoTengoUnLCD. I edited it. – Friedrich Philipp Mar 04 '16 at 17:17

How to solve $2f-f_u=g(v)$?

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