If dy/dt = 2y/(t(t+2)) for t>0 and y = 1 when t = 1, then when t = 2, y = ?
To find the value of y when t = 2, we need to solve the differential equation dy/dt = 2y/(t(t+2)) and use the initial condition y = 1 when t = 1.
First, let's separate the variables by multiplying both sides of the equation by (t(t+2)):
(t(t+2)) dy/dt = 2y
Next, let's move the y term to one side and the t terms to the other side:
dy/y = 2 dt / (t(t+2))
Now, we can integrate both sides with respect to their respective variables. For the left side, we integrate 1/y with respect to y, and for the right side, we integrate 2 dt / (t(t+2)) with respect to t:
∫(1/y) dy = ∫(2 dt / (t(t+2)))
The left side can be integrated as the natural logarithm of y:
ln(y) = ∫(2 dt / (t(t+2)))
Let's calculate the integral on the right side:
ln(y) = ∫(2 dt / (t^2 + 2t))
We can apply partial fraction decomposition to simplify the integral:
ln(y) = ∫(A/t + B/(t+2)) dt
To find the values of A and B, we can multiply both sides of the equation by the denominator (t(t+2)):
(t(t+2)) ln(y) = A(t+2) + Bt
Now, we can set t = 0 to eliminate one of the variables:
(0(0+2)) ln(y) = A(0+2) + B(0)
0 = 2A
Therefore, A = 0.
Now, substitute A = 0 back into the equation:
(t(t+2)) ln(y) = 0 + Bt
(t(t+2)) ln(y) = Bt
Since this equation holds for all values of t, we can equate the coefficients on both sides:
ln(y) = B
Now, let's find the value of B by substituting t = 1 and y = 1:
ln(1) = B
0 = B
Therefore, B = 0.
Now, substitute B = 0 back into the equation:
ln(y) = 0
Therefore, taking the exponential of both sides gives:
y = e^0
y = 1
So when t = 2, y = 1.