Solve the following initial value problem and determine the interval where the solution exists.
dy/dt = 2ty^2
y(0) = y_0
Textbook answer is:
When y_0 <> 0, then y = 1/(1/y_0 - t^2).
When y_0 = 0, then y = 0
When y_0 > 0, the interval where solution exists is |t| < 1/sqrt(y_0)
When y_0 <= 0, the interval where solution exists is -infinity < t < +infinity
My work:
The equation is separable:
dy/y^2 = 2t dt
-1/y = t^2 + C
When y_0 = 0, I use intuition to come to the same solution that the textbook shows.
When y_0 <> 0
C = -1/y_0
y = 1/(1/y_0 - t^2).
The interval where the solution exists is where the denominator expression does not equal zero. Algebraically, that simplifies to t <> 1/sqrt(y_0)
The textbook answer is different. What did I do wrong or omit?
Your approach is correct, but there seems to be a slight confusion in your interpretation of the result.
You correctly separated the variables and obtained the equation:
-1/y = t^2 + C
When y_0 = 0, you correctly interpreted this as the trivial solution y = 0, since the denominator becomes infinite.
When y_0 ≠ 0, you solved for y:
y = 1/(1/y_0 - t^2)
Now, to determine the interval where the solution exists, we need to consider the denominator expression. You correctly identified that the denominator should not equal zero.
The denominator is 1/y_0 - t^2. To find where this expression is defined, we set it equal to zero and solve:
1/y_0 - t^2 = 0
t^2 = 1/y_0
t = ± √(1/y_0)
Now, let's consider the two cases separately:
1) When y_0 > 0:
In this case, 1/y_0 > 0, and therefore, t can be any real number. Thus, the solution exists for all t, which means the interval where the solution exists is -∞ < t < +∞.
2) When y_0 ≤ 0:
In this case, 1/y_0 ≤ 0, and therefore, the square root √(1/y_0) is not defined. Since the denominator cannot be zero, the solution is not defined for any value of t. Thus, there is no interval where the solution exists.
Therefore, the textbook answer is correct:
- When y_0 > 0, the interval where the solution exists is -∞ < t < +∞.
- When y_0 ≤ 0, the solution is not defined for any value of t.