Uppgift 6

Kedjeregeln ger:
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u},$$
$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}=3y^2\frac{\partial f}{\partial u}+\frac{\partial f}{\partial v},$$
eftersom
$$\frac{\partial u}{\partial x}=1,\quad\frac{\partial v}{\partial x}=0,\quad\frac{\partial u}{\partial y}=3y^2,\quad\text{och}\quad\frac{\partial v}{\partial y}=1.$$
Insatt i differentialekvationen ger detta
$$0=3y^2\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}=3y^2\frac{\partial f}{\partial u}-3y^2\frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}=-\frac{\partial f}{\partial v},$$
dvs i variablerna $u,v$ får vi ekvationen $\frac{\partial f}{\partial v}=0$, som innebär att $f$ inte beror på $v$, så vi kan skriva $f=h(u)$. I de ursprungliga variablerna $x,y$ får vi alltså att $f(x,y)=h(x+y^3)$. Bivillkoret ger nu att
$$\sin x=f(x,0)=h(x),$$
dvs $h(x)=\sin x$, och $f(x,y)=\sin(x+y^3)$.