Question

dada a equaçao diferencial de 2 ordem

Answer 1

Resposta:

a.

Explicação passo a passo:

$f''(x)=3\sin x-4\cos x\iff f'(x)=\int(3\sin x-4\cos x)\;dx$

$f'(x)=-3\cos x-4\sin x+C$

Sendo $f'(0)=2$:

$-3\cos 0-4\sin0+C=2$

$-3\cdot1-4\cdot0+C=2\iff C=5$

Concluindo assim que $f'(x)=-3\cos x-4\sin x+5$.

$f'(x)=-3\cos x-4\sin x\iff f(x)=\int(-3\cos x-4\sin x+5)\;dx$

$f(x)=-3\sin x+4\cos x+5x+C$

Sendo $f(0)=7$:

$-3\sin 0+4\cos 0+5\cdot0+C=7$

$-3\cdot0+4\cdot1+C=7\iff C=3$

Concluindo assim que $f(x)=-3\sin x+4\cos x+5x+3$.