if dy/dx=-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 when y=1/3, we can solve the differential equation dy/dx = -3y.
The given differential equation dy/dx = -3y is a separable differential equation. We can rearrange it as:
1/y dy = -3 dx
Next, we can integrate both sides with respect to their respective variables:
∫(1/y) dy = -3 ∫dx
ln|y| = -3x + C
where C is the constant of integration.
Now, we can solve for y by taking the exponential of both sides:
e^(ln|y|) = e^(-3x+C)
|y| = e^(-3x) * e^C
Since e^C is a positive constant, we can simplify as:
y = Ae^(-3x)
where A = e^C is a positive constant.
Using the given initial condition y=1 when t=0, we can substitute these values into the equation:
1 = Ae^(-3*0)
1 = Ae^0
1 = A
So, A = 1.
Now, we can rewrite the equation as:
y = e^(-3x)
To find the value of t when y=1/3, we can substitute y=1/3 into the equation:
1/3 = e^(-3x)
To solve for x, we can take the natural logarithm of both sides:
ln(1/3) = -3x
ln(1/3) / -3 = x
Now, we have the value of x when y=1/3. However, the question asks for the value of t. Since t is given as 0, we can use the relation t = x and find the value of t as:
t = ln(1/3) / -3
Thus, the value of t for which y=1/3 is ln(1/3) / -3.