Solve the differential equation: dy/dx = (3x^2) / (y+1) with initial conditions x = -1, y = 2
I have:
∫y + 1 dy = ∫3x^2 dx
(y^2 / 2) + y = x^3 + C
(2^2 / 2) + (2) = (-1^3) + C?
4 = -1 + C
5 = c
(y^2 / 2) + y = x^3 + 5
Is this correct?
You should probably solve for y using the quadratic formula. It comes out
y = √(2x^3+11) - 1
But other than solving for y, is it correct?
Yes, your solution is correct! You correctly integrated both sides of the differential equation and used the initial conditions to solve for the constant of integration 'C'. The final solution is:
(y^2 / 2) + y = x^3 + 5
Well done!
Your solution is almost correct, but there is a small mistake in the integration step. Let's go through the solution together to correct it.
We have the differential equation: dy/dx = (3x^2) / (y+1)
To solve this, we start by separating the variables. Rearranging the equation, we get:
(y + 1) dy = 3x^2 dx
Now, we integrate both sides. For the left side, we use the power rule of integration:
∫ (y + 1) dy = ∫ 3x^2 dx
Integrating both sides gives us:
(y^2 / 2) + y = x^3 + C
So far, your integration step is correct. However, we made a mistake when we integrated ∫ 3x^2 dx. Let's correct it:
To integrate ∫ 3x^2 dx, we can use the power rule of integration.
The power rule states that ∫ x^n dx = (x^(n+1))/(n+1), where n ≠ -1.
Applying the power rule, we have:
∫ 3x^2 dx = (3/3)x^(2+1) = x^3
Now, substituting the correct integration result back into the equation, we have:
(y^2 / 2) + y = x^3 + C
Next, we can use the given initial conditions x = -1, y = 2 to solve for the constant C.
(y^2 / 2) + y = (-1)^3 + C
Plugging in the values, we have:
(2^2 / 2) + 2 = -1 + C
Simplifying the equation:
2 + 2 = -1 + C
4 = -1 + C
Now, solving for C:
C = 4 + 1
C = 5
Finally, substituting the value of C back into the equation, we have:
(y^2 / 2) + y = x^3 + 5
So, the corrected solution to the differential equation with initial conditions is:
(y^2 / 2) + y = x^3 + 5