Calcule C1 e C2 de modo que y(x)=C1senx+C2cosx satisfaça as condições dadas: y(0)=2; y'(0)=1.

Seja :

\(y(x)=C1senx+C2cosx\)

Temos:

\(y(x)=C1senx+C2cosx\\ y(0)=C1sen0+C2cos0\\ y(0)=C.0+C2.1\\ y(0)=C2\\ 2=C2\\ C2=2 \)

\(y(x)=C1senx+C2cosx\\ y'(x)=C1.cosx+C2(-senx)\\ y'(x)=C1.cosx-C2.senx)\\ y'(0)=C1.cos0-C2.sen.0)\\ y'(0)=C1.1-C2.0)\\ y'(0)=C1)\\ C1=1\)

Assim

\(\boxed{C1=1 \\C2=2}\)

y'(x)= c1·cosx - c2·sen(x)