What is the solution of the differential equation of (6x+1)y^2 dy/dx + 3x^2 +2y^3=0?
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To find the solution of the given differential equation, we can use the method of separation of variables.
Step 1: Rearrange the equation:
To begin, let's rearrange the equation to separate the variables. Notice that we have both dy and dx terms, so we need to group them together.
(6x+1)y^2 dy + (3x^2 + 2y^3) dx = 0
Step 2: Divide by y^2:
To separate the variables, divide both sides of the equation by y^2.
(6x+1) dy + (3x^2 + 2y^3) dx / y^2 = 0
Step 3: Rearrange terms:
Next, let's rearrange the equation to isolate dy and dx terms on each side.
(6x+1) dy = - (3x^2 + 2y^3) dx / y^2
Step 4: Cross-multiply:
To further separate the variables, we can cross-multiply.
(6x+1) y^2 dy = - (3x^2 + 2y^3) dx
Step 5: Integrate both sides:
Now, we integrate both sides of the equation.
∫ (6x+1) y^2 dy = ∫ -(3x^2 + 2y^3) dx
The integration on the left side depends on y, so we treat x as a constant. Similarly, the integration on the right side depends on x, so we treat y as a constant.
Integrating the left side:
∫ y^2 dy = ∫ (-(3x^2 + 2y^3)) dx
By integrating the left side, we get:
(1/3) y^3 = - (x^3 + 2y^3x + C)
Note that C represents the constant of integration.
Step 6: Solve for y:
To find the solution, we solve the equation for y.
y^3 = -3(x^3 + 2y^3x + C)
y^3 + 6xy^3 + 3x^3 + 3C = 0
This equation represents the solution to the given differential equation.