Resolva a equação diferencial

Prévia do material em texto

Resolva a equação diferencial y′′−2y′=sen(4x)y″−2y′=sen(4x) com y(0)=140y(0)=140 e y′(0)=95y′(0)=95.
		
	 
	y=e2x−1+140cos4x−120sen(4x)y=e2x−1+140cos4x−120sen(4x)
	
	y=e2x−1+120cos4x−140sen(4x)y=e2x−1+120cos4x−140sen(4x)
	
	y=1+e2x+120cos4x−120sen(4x)y=1+e2x+120cos4x−120sen(4x)
	
	y=1+e2x−140cos4x+120sen(4x)y=1+e2x−140cos4x+120sen(4x)
	
	y=1−e2x−140cos4x−120sen(4x)y=1−e2x−140cos4x−120sen(4x)
	Respondido em 17/11/2021 16:50:46
	
	Explicação:
A resposta correta é: y=e2x−1+140cos4x−120sen(4x)y=e2x−1+140cos4x−120sen(4x)