Please help with this very complicated question! Using 3(x-3)(x^2-6x+23)^1/2, as the chain rule fifferentiation of f(x)=(x^2-6x+23). Please explain how I find the general solution to 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'll follow these steps:
Step 1: Separate the variables.
Step 2: Integrate both sides.
Step 3: Rearrange the equation to solve for y.
Let's go through each step in detail:
Step 1: Separate the variables
The given differential equation is dy/dx = (2/27)*(x-3)*√(x^2-6x+23)/y.
To separate the variables, we'll multiply both sides by y and divide both sides by √(x^2-6x+23):
y * dy = (2/27)*(x-3) * √(x^2-6x+23) * dx / √(x^2-6x+23)
Step 2: Integrate both sides
Now, we'll integrate both sides of the equation.
∫ y * dy = ∫ (2/27)*(x-3) * dx
On the left-hand side, we integrate y with respect to y, giving us (1/2)*y^2. On the right-hand side, we integrate (2/27)*(x-3) with respect to x, giving us (1/27)*(x-3)^2 + C (where C is the constant of integration).
So, the equation becomes:
(1/2)*y^2 = (1/27)*(x-3)^2 + C
Step 3: Solve for y
Now we rearrange the equation to solve for y:
Multiply both sides by 2:
y^2 = (2/27)*(x-3)^2 + 2C
Take the square root of both sides:
y = ± √((2/27)*(x-3)^2 + 2C)
Note that we take the ± because we have to account for both the positive and negative square roots.
So, the general solution to the given differential equation, in implicit form, is:
y = ± √((2/27)*(x-3)^2 + 2C)
where C is the constant of integration.