if dy/dt=-3y and if y=1 when t=0, what is the value of t for which y=1/3?
To find the value of t for which y is equal to 1/3, we need to solve the given differential equation:
dy/dt = -3y
First, let's separate the variables by multiplying both sides of the equation by dt and dividing by y:
1/y dy = -3 dt
Now, let's integrate both sides of the equation:
∫(1/y) dy = ∫(-3) dt
ln|y| = -3t + C
Where C is the constant of integration.
Next, let's apply the initial condition y = 1 when t = 0. Plugging these values into the equation:
ln|1| = -3(0) + C
ln|1| = 0 + C
ln(1) = C
Since the natural logarithm of 1 is 0, C = 0.
Therefore, the equation becomes:
ln|y| = -3t
To solve for t when y = 1/3, we substitute y = 1/3 into the equation:
ln|1/3| = -3t
-ln(3) = -3t
Now, we can solve for t by dividing both sides by -3:
t = ln(3)/3
Hence, the value of t for which y is equal to 1/3 is t = ln(3)/3.