I posted this below, and no one answered. Please help with this very complicated question! Using 3(x-3)(x^2-6x+23)^1/2, as the chain rule differentiation of f(x)=(x^2-6x+23). Please explain how I find the general solution to the differential equation dy/dx= 2/27(x-3)SQUARE ROOT BEGINS (x^2-6x+23)/y SQUARE ROOT ENDS (y>0), giving answer in implicit form. I need details! Many thanks.
To find the general solution to the given differential equation, we can start by rewriting it in a more standard form.
The differential equation is:
dy/dx = (2/27)(x-3)√(x^2-6x+23)/y, where y > 0.
Step 1: Separate variables.
To separate the variables, we multiply both sides of the equation by y and divide by √(x^2-6x+23):
ydy = (2/27)(x-3)dx.
Step 2: Integrate both sides.
Now we integrate both sides of the equation with respect to their respective variables:
∫ydy = ∫(2/27)(x-3)dx.
To integrate ydy, we use the power rule for integration: ∫ydy = (1/2)y^2 + C1.
To integrate (2/27)(x-3)dx, we use the power rule for integration: ∫(2/27)(x-3)dx = (1/27)(x-3)^2 + C2.
Therefore, our equation becomes:
(1/2)y^2 + C1 = (1/27)(x-3)^2 + C2.
Step 3: Combine the constants.
We can combine the constants into a single constant, C:
(1/2)y^2 = (1/27)(x-3)^2 + C.
Step 4: Multiply by 2.
To get rid of the fraction, we multiply both sides of the equation by 2:
y^2 = (2/27)(x-3)^2 + 2C.
Step 5: Take the square root.
To solve for y, we take the square root of both sides:
y = √((2/27)(x-3)^2 + 2C).
This is the implicit form of the general solution to the given differential equation.