Chapter 1 - First-Order Differential Equations - 1.2 Basic Ideas and Terminology - True-False Review - Page 21: d

True

Obtain: $y(x)=c_1\cos(x)+5c_2\sin(x)\\y'(x)=c_1\sin(x)-5c_2\cos(x)\\y''(x)=-c_1\cos (x)-5c_2\sin(x)$ We will substitute these equations in the differential equation: $-c_1\cos(x)-5c_2\sin(x)+c_1\cos(x)+5c_2\sin(x)=0\\ \Leftrightarrow 0=0$ Hence, the general solution to $y′′ + y = 0$ is $y(x) =c_1 \cos x + 5c_1 \sin x$