Find general solution.
2ty dy/dt = 3y^2 - t^2
use separation of variables to get
y = ±x√(cx+1)
To find the general solution of the differential equation 2ty(dy/dt) = 3y^2 - t^2, we can follow these steps:
Step 1: Rewrite the equation in a standard form.
Start by rearranging the equation:
2ty(dy/dt) = 3y^2 - t^2
Step 2: Separate the variables.
To do this, divide both sides of the equation by (3y^2 - t^2):
(2ty)/(3y^2 - t^2) dy/dt = 1
Step 3: Integrate both sides with respect to t.
Now we integrate both sides with respect to t:
∫(2ty)/(3y^2 - t^2) dy/dt dt = ∫1 dt
Step 4: Simplify the left-hand side by using substitution.
Let's substitute u = 3y^2 - t^2, which means du/dt = -2t. Therefore, dt = (-1/2) * du/t.
The equation becomes:
∫(2y)/(u) * (-1/2) * du = ∫1 dt
Step 5: Integrate both sides.
Now we integrate both sides:
(-1/2) ∫(2y)/(u) du = ∫1 dt
Step 6: Simplify the left-hand side.
The integral on the left-hand side simplifies to:
(-1/2) ln|u| + C1
Step 7: Integrate the right-hand side.
The integral on the right-hand side is simply:
∫1 dt = t + C2
Step 8: Combine the results.
Combining the two sides gives us the general solution of the differential equation:
(-1/2) ln|u| = t + C2
Step 9: Replace u with its original expression.
We recall that u = 3y^2 - t^2, so we substitute it back in:
(-1/2) ln|3y^2 - t^2| = t + C2
Step 10: Simplify further, if desired.
If you want to simplify the equation further, you can use properties of logarithms to write it as:
ln|3y^2 - t^2| = -2t - 2C2
This is the general solution to the given differential equation.
To find the general solution of the given differential equation, we can use the method of separation of variables.
First, let's rewrite the equation in a more familiar form:
2ty dy = (3y^2 - t^2) dt
Now, we can separate the variables by dividing both sides of the equation:
(2y dy) / (3y^2 - t^2) = t dt
Next, we integrate both sides with respect to their respective variables. Starting with the left side:
∫ (2y dy) / (3y^2 - t^2) = ∫ t dt
To integrate the left side, we can use the method of partial fractions. We need to factor the denominator:
3y^2 - t^2 = (√3y + t)(-√3y + t)
We can rewrite the left side as follows:
∫ (2y dy) / [(√3y + t)(-√3y + t)] = ∫ t dt
Using partial fractions, we can express the integrand on the left side as:
A / (√3y + t) + B / (-√3y + t)
To determine the values of A and B, we can multiply both sides of the equation by the denominator, then rearrange the terms:
2y = A(-√3y + t) + B(√3y + t)
Expanding and collecting like terms:
2y = (-A√3 + B√3)y + (At + Bt)
Comparing the coefficients of y and the constant terms separately, we get two equations:
-√3A + √3B = 2 (1)
A + B = 0 (2)
From equation (2), we can express A in terms of B:
A = -B
Substituting this into equation (1), we get:
-√3(-B) + √3B = 2
√3B + √3B = 2
2√3B = 2
B = 1/√3
Substituting the value of B back into equation (2), we can solve for A:
A + 1/√3 = 0
A = -1/√3
Now that we have found the values of A and B, we can replace the integrand on the left side with the partial fraction decomposition:
∫ (2y dy) / [(√3y + t)(-√3y + t)] = ∫ (-1/√3) / (√3y + t) + (1/√3) / (-√3y + t) dt
Integrating each term separately:
(-1/√3) ∫ 1 / (√3y + t) dy + (1/√3) ∫ 1 / (-√3y + t) dy = ∫ t dt
The integral of 1 / (√3y + t) can be evaluated using a logarithmic function:
(-1/√3) ln |√3y + t| + (1/√3) ln |√3y - t| = ∫ t dt + C
Simplifying the right side integral:
(-1/√3) ln |√3y + t| + (1/√3) ln |√3y - t| = (1/2) t^2 + C
Combining the logarithmic terms:
(-1/√3) ln |(√3y + t) / (√3y - t)| = (1/2) t^2 + C
Taking the exponential of both sides to eliminate the natural logarithm:
|(√3y + t) / (√3y - t)| = e^((1/2) t^2 + C)
Removing the absolute value sign:
(√3y + t) / (√3y - t) = ± e^((1/2) t^2 + C)
Now, we can solve for y by cross-multiplying:
(√3y + t) = ± e^((1/2) t^2 + C) (√3y - t)
Expanding and rearranging the terms for y:
√3y + t = ± (√3y - t) e^((1/2) t^2 + C)
Distributing the exponential term on the right:
√3y + t = ± (√3e^C)(√3y - t) e^(1/2 t^2)
Expanding further:
√3y + t = ± (√3e^C)(√3y - t) e^(1/2 t^2)
Simplifying:
√3y + t = ± 3y - t e^(1/2 t^2)
Rearranging terms:
(√3y - 3y) = (-t - t e^(1/2 t^2))
Combining like terms:
(-2y) = (-t(1 + e^(1/2 t^2)))
Finally, we can divide both sides by -2 to isolate y:
y = t(1 + e^(1/2 t^2)) / 2
Therefore, the general solution to the given differential equation is:
y = t(1 + e^(1/2 t^2)) / 2