Solve the following separable equation.
y'=(4x^3-1)/y for y(1)= -3
y dy=4x^3 dx-dx
1/2 y^2=x^4-x + C
y= sqrt(2x^2-2x) + C1
y(1)=sqrt(0)===> C1=-3
y=sqrt(2x^3-2x) -3
check this
dy/dx * y = 4x^3 -1
integral of y dy = integral of (4x^3-1)dx
y^2/2 = x^4 - x + C
9/2 = 1 -1 + C
C = 9/2
y^2 = 2x^4 -2x +9
y = -sqrt(2x^4 -x +9)
To solve the given separable equation, y' = (4x^3 - 1)/y, we can follow these steps:
Step 1: Separate the variables
Move the y term to one side and the x term to the other side:
y dy = (4x^3 - 1) dx
Step 2: Integrate both sides
Integrate both sides of the equation with respect to their respective variables:
∫y dy = ∫(4x^3 - 1) dx
∫y dy can be simplified to (1/2)y^2, and ∫(4x^3 - 1) dx can be expanded to x^4 - x:
(1/2)y^2 = ∫(4x^3) dx - ∫(1) dx
(1/2)y^2 = x^4 - x + C1
Here, C1 is the constant of integration.
Step 3: Solve for y
To solve for y, we can take the square root of both sides:
y = ±√(2x^4 - 2x + 2C1)
Step 4: Apply the initial condition
Using the given initial condition y(1) = -3, we can substitute x = 1 and y = -3 into the equation and solve for C1.
-3 = ±√(2(1)^4 - 2(1) + 2C1)
-3 = ±√(2 - 2 + 2C1)
-3 = ±√(2C1)
Since we know that y(1) = -3, we can disregard the positive square root solution and take the negative square root solution:
-3 = -√(2C1)
Squaring both sides:
9 = 2C1
C1 = 9/2
Step 5: Substitute C1 into the general solution
Now that we know the value of C1, we can substitute it back into the general solution equation:
y = ±√(2x^4 - 2x + 2(9/2))
y = ±√(2x^4 - 2x + 9)
Therefore, the solution to the separable equation y' = (4x^3 - 1)/y with the initial condition y(1) = -3 is y = ±√(2x^4 - 2x + 9).