$Equation with partial derivative$

Question

$Equation with partial derivative$

Find all the functions f that are continuously differentiable such that:
$\frac{\partial f}{\partial x}-3\frac{\partial f}{\partial y}=0 $

We were given a beneficial hint:
define: u(x,y)= ax + by
v(x.y)=cx+dy
f(x,y)=F(u(x,y),v(x,y))
ad-bc≠0

Multivariable Calculus

Danielatia123

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Let $u=-(x+y)/2$ and $v=(3x+y)/2$. Then $x=u+v$ and $y=-3u-v$. Let $$F(u,v)=f(x(u,v),y(u,v))=f(u+v,-3u-v) $$ Then we have $$f(x,y)=F(u(x,y),v(x,y))=F(-(x+y)/2,(3x+y)/2) $$ Now we have $$\partial _u F(u,v)=(\partial x/\partial u)\partial_xf +(\partial y/\partial u)\partial_yf=\partial_xf -3\partial_yf =0$$ So $F$ does not depend on $u$ and we have $F(u,v)=g(v)$ for some function $g$. Therefore $$ f(x,y)=F(u(x,y),v(x,y))=g(v(x,y))=g((3x+y)/2)$$for an arbitrary continuously differentiable function of one variable $g$. This is the general solution and can be easily checked that it satisfies the equation. Since $g(t)$ is a function of one variable for $t=(3x+y)/2$ we have $$\partial _x f =g'(t)\partial_x t=3g'(t)/2$$ Similarly $\partial_y f=g'(t)/2$. Hence $$\partial_x f-3\partial_yf=3g'(t)/2-3g'(t)/2=0$$

Martin

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Danielatia123

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I'm not sure how you differentiate f(x,y) if f(x,y)=g((3x+y)/2), in which g is a single variable. Can you show it, please?
- Martin
  
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  I added the explanation at the end.

$Equation with partial derivative$

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