If dy/dt=-2y and if y=1 when t=0 what is the value of t for which y=1/2 i (A) -(ln2)/(2) (B) -(1)/(4) (C) (ln2)/(2) (D) (sqrt (2))/(2) (E) ln2

To solve the differential equation $\frac{dy}{dt} = -2y$ with the initial condition $y(0) = 1$, we can use the method of separation of variables.<br /><br />First, we separate the variables by dividing both sides of the equation by $y$ and multiplying both sides by $dt$:<br /><br />$\frac{1}{y} \, dy = -2 \, dt$<br /><br />Now, we can integrate both sides:<br /><br />$\int \frac{1}{y} \, dy = \int -2 \, dt$<br /><br />$\ln|y| = -2t + C$<br /><br />where $C$ is the constant of integration.<br /><br />To find the value of $t$ for which $y = \frac{1}{2}$, we substitute $y = \frac{1}{2}$ into the equation:<br /><br />$\ln\left|\frac{1}{2}\right| = -2t + Cln\left(\frac{1}{2}\right) = -2t + C$<br /><br />Since $\ln\left(\frac{1}{2}\right) = -\ln(2)$, we have:<br /><br />$-\ln(2) = -2t + C$<br /><br />Now, we can solve for $t$:<br /><br />$-\ln(2) + \ln(2) = -2t + C + \ln(2)$<br /><br />$0 = -2t + C + \ln(2)$<br /><br />$2t = C + \ln(2)$<br /><br />$t = \frac{C + \ln(2)}{2}$<br /><br />Since $y(0) = 1$, we can substitute $t = 0$ into the equation $\ln|y| = -2t + C$ to find the value of $C$:<br /><br />$\ln|1| = -2(0) + C$<br /><br />$0 = C$<br /><br />Therefore, $t = \frac{0 + \ln(2)}{2} = \frac{\ln(2)}{2}$.<br /><br />So, the correct answer is (C) $\frac{\ln(2)}{2}$.