If dy/dx = xy - y^2 and y(1) = 3, then y (2) = ?
Well, let's see if I can solve this equation without clowning around too much.
To solve the differential equation, we can use separation of variables. We start by dividing both sides by (xy - y^2):
dy/dx / (xy - y^2) = 1
Now, we can rewrite the left side as a partial fraction:
dy/((x - y)y) = 1
Splitting the fraction, we get:
(y / (x - y)) dy = dx
Integrating both sides, we have:
∫ (y / (x - y)) dy = ∫ dx
Integrating the left side can be a bit tricky, but with a little algebraic manipulation, we get:
-∫ (y / (y - x)) dy = ∫ dx
Now, we can integrate using substitution. Let u = y - x, which implies du = dy - dx. Substituting these values, we get:
-∫ (1 / u) du = ∫ dx
Applying the limits, we have:
-ln|u| = x + C
Now, let's solve for u:
-ln|u| = x + C
|u| = e^(-x - C)
We know that u = y - x, so we can rewrite it as:
|y - x| = e^(-x - C)
Since we have two cases, we need to deal with the absolute value. Let's consider each case separately:
Case 1: y - x > 0
In this case, we can remove the absolute value:
y - x = e^(-x - C)
Now, we can solve for y:
y = e^(-x - C) + x
Now let's apply the initial condition y(1) = 3:
3 = e^(-1 - C) + 1
2 = e^(-1 - C)
-ln(2) = -1 - C
C = ln(2) + 1
Now we can find y(2):
y(2) = e^(-2 - ln(2) - 1) + 2
y(2) = e^(-ln(2) - 3)
y(2) = 2^(-1) * e^(-3)
y(2) = 1/2 * e^(-3)
Now that we've solved the equation, I hope you find it more amusing than confusing!+
To find the value of y(2), we need to solve the given differential equation using the initial condition y(1) = 3 and then substitute x = 2 into the solution.
Given: dy/dx = xy - y^2 and y(1) = 3.
Separating variables, we have:
dy / (y - y^2) = x dx
Now, integrate both sides of the equation:
∫(1 / (y - y^2)) dy = ∫x dx
To integrate the left-hand side, we can use partial fraction decomposition:
1 / (y - y^2) = A / y + B / (1 - y)
Multiplying through by (y - y^2), we get:
1 = A(1 - y) + By
Expanding and matching coefficients, we find:
A = 1
B = 1
Therefore, the integral becomes:
∫(1/y) dy + ∫(1 / (1 - y)) dy = ∫x dx
ln |y| - ln |1 - y| = (1/2) x^2 + C1
Taking the exponential of both sides:
|y| / |1 - y| = e^[(1/2) x^2 + C1]
Removing the absolute value signs, we get:
y / (1 - y) = Ke^[(1/2) x^2]
where K is a constant.
Now, substitute the initial condition y(1) = 3:
3 / (1 - 3) = K e^(1/2)
-3 = K e^(1/2)
Solving for K, we find:
K = -3 / e^(1/2)
So, the particular solution is:
y / (1 - y) = (-3 / e^(1/2)) e^(1/2) x^2
Simplifying:
y / (1 - y) = -3 x^2
Multiplying both sides by (1 - y):
y = -3 x^2 (1 - y)
Expanding:
y = -3 x^2 + 3 xy
Now we can substitute x = 2 into this equation to find y(2):
y(2) = -3(2)^2 + 3(2)(y)
Simplifying:
y(2) = -3(4) + 6y
y(2) = -12 + 6y
To find the value of y(2), we need more information about the function y(x) or another equation with another initial condition. Without further information, we cannot find the exact value of y(2).
To find the value of y at x = 2, we can use the given initial condition y(1) = 3 and the differential equation dy/dx = xy - y^2.
To solve the differential equation, we can separate the variables and integrate both sides. Here's how:
dy/(y-y^2) = x dx
First, factor out y from the denominator:
dy/y(1-y) = x dx
Now, we can split this into partial fractions:
A/y + B/(1-y) = dy/y(1-y)
To find the values of A and B, we can multiply both sides by the common denominator y(1-y):
A(1-y) + By = 1
Expanding the left side:
A - Ay + By = 1
From this equation, we can equate the coefficients of y to find A and B:
-A + B = 0 (coefficient of y on the left side)
A = 1 (coefficient of y on the right side)
From the first equation, we have B = A = 1.
So, our integral now becomes:
∫[A/y + B/(1-y)] dy = ∫(1) dx
Integrating both sides:
A ln|y| - B ln|1-y| = x + C
Plugging in the values A = B = 1:
ln|y| - ln|1-y| = x + C
Combining the logarithms:
ln|y/(1-y)| = x + C
Exponentiating both sides:
|y/(1-y)| = e^(x + C)
Since the absolute value can be positive or negative, we'll consider the positive and negative cases separately.
Positive case: y/(1-y) = e^(x + C)
Simplifying this equation:
y = (1 - y)e^(x + C)
Expanding:
y = e^(x + C) - ye^(x + C)
Bringing the y terms to one side:
y + ye^(x + C) = e^(x + C)
Factoring out y:
y(1 + e^(x + C)) = e^(x + C)
Dividing both sides by (1 + e^(x + C)):
y = e^(x + C) / (1 + e^(x + C))
Using the initial condition, y(1) = 3, we can find the value of C:
3 = e^(1 + C) / (1 + e^(1 + C))
Multiply both sides by (1 + e^(1 + C)):
3(1 + e^(1 + C)) = e^(1 + C)
Expanding:
3 + 3e^(1 + C) = e^(1 + C)
Bringing all terms to one side:
e^(1 + C) - 3e^(1 + C) - 3 = 0
Now, we can substitute z = e^(1 + C) to get a quadratic equation:
z - 3z - 3 = 0
Simplifying:
-2z - 3 = 0
Solving for z:
z = -3/2
Substituting z back in:
e^(1 + C) = -3/2
Since e^(x) is always positive, there are no real solutions for this equation. So, the positive case does not yield a valid solution.
Negative case: y/(1-y) = -e^(x + C)
Using the same steps as before, we find:
y = e^(x + C) / (e^(x + C) + 1)
Using the initial condition y(1) = 3, we can find the value of C:
3 = e^(1 + C) / (e^(1 + C) + 1)
Multiply both sides by (e^(1 + C) + 1):
3 (e^(1 + C) + 1) = e^(1 + C)
Expanding:
3e^(1 + C) + 3 = e^(1 + C)
Bringing all terms to one side:
e^(1 + C) - 3e^(1 + C) - 3 = 0
Now, we can substitute z = e^(1 + C) to get a quadratic equation:
z - 3z - 3 = 0
Simplifying:
-2z - 3 = 0
Solving for z:
z = -3/2
Substituting z back in:
e^(1 + C) = -3/2
Since e^(x) is always positive, there are no real solutions for this equation. So, the negative case does not yield a valid solution either.
Therefore, there is no valid solution for y(2) using the given differential equation dy/dx = xy - y^2 and the initial condition y(1) = 3.