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$f(tx,ty)=t^5f(x,y)$ for all values of $x, y, t$ where both functions are differentiable. Show that $$a)\ xf_x+yf_y=5f$$ $$b)\ \ x^2f_{xx}+2xyf_{xy}+y^2f_{yy}=20f$$

Clearly, there is differentiating of the initial equation in order to get to a. And a double differentiation to get to b. But, I am confused at differentiating it. I think it needs the chain rule but the composition of the different variables is confusing me.

a) $$\frac{\partial f(tx, ty)}{\partial t}=f_x*\frac{d(tx)}{dt}+f_y*\frac{d(ty)}{dt}=xf_x+yf_y$$

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  • $\begingroup$ The first equaality is Euler's formula for the homogeneous functions of degree $5$. $\endgroup$ – daulomb Nov 11 '13 at 15:42
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In order to get the first equality differentiate the equality $f(tx, ty)=t^5f(x, y)$ with respect to $t$ and then put $t=1$. For the second one differentiate w.r. t. $t$ twice and put $t=1$. Differentiate w.r.t. Let $u=tx$ and $v=ty$. Then differentiating either side of the inequality w.r.t. $t$ along with the use of chain rule gives: $f_u u'(t)+f_v v'(t)=5t^4f(x, y)=xf_x+yf_y=5f$. Notice that we put $t=1$ in the last equality and $u'(t)=x, v'(t)=y$. Differentiating $f_u u'(t)+f_v v'(t)=5t^4f(x, y)=xf_x+yf_y=5f$ w.r.t. $t$ we have $\big(f_{uu}u'(t)+f_{uv}v'(t)\big)x+\big(f_{vu}u'(t)+f_{vv}v'(t)\big)y=20t^3f(x, y)$. For $t=1$ you get the second equality. Observe that when $t=1$ you have $f_{uu}=f_{xx}$, $f_{uv}=f_{xy}=f_{yx}$ and $f_{vv}=f_{yy}$ and remember $u'(t)=x$ and $v'(t)=y$.

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  • $\begingroup$ I made a few additions to the question to clarify... $\endgroup$ – John Nov 11 '13 at 15:51
  • $\begingroup$ As for the second one, can you show me how its done. I cant see how $f_{xy}$ came into the equation $\endgroup$ – John Nov 11 '13 at 15:53
  • $\begingroup$ Have you got it? $\endgroup$ – daulomb Nov 11 '13 at 16:01
  • $\begingroup$ I think I got it. Thanks $\endgroup$ – John Nov 11 '13 at 16:17

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