Find the solution of the initial value problems $\frac{d K}{d t}=2-\cos 5 t$ when $K(0)=-12$ .
A) $2 t-\frac{\sin 5 t}{5}-12$
B) $2 t-\frac{\sin 5 t}{5}+12$
C) $2 t+\frac{\sin 5 t}{5}-12$
D) $2 t+\frac{\sin 5 t}{5}+12$

Question

Quizplus · Accepted Answer

To solve the initial value problem $\frac{d K}{d t}=2-\cos 5 t$ with $K(0)=-12$, we integrate the right-hand side with respect to $t$. The integral of $2$ with respect to $t$ is $2t$, and the integral of $-\cos 5t$ with respect to $t$ is $-\frac{\sin 5t}{5}$. Thus, the general solution is $K(t) = 2t - \frac{\sin 5t}{5} + C$, where $C$ is the constant of integration. To find $C$, we use the initial condition $K(0) = -12$, which gives us $C = -12$ when we substitute $t = 0$. Therefore, the solution is $K(t) = 2t - \frac{\sin 5t}{5} - 12$.

Find the Solution of the Initial Value Problems dKdt=2−cos⁡5t\frac{d K}{d t}=2-\cos 5 tdtdK​=2−cos5t

Find the Solution of the Initial Value Problems $\frac{d K}{d t}=2-\cos 5 t$