Find for dy/dx 4 – xy = y^3.
What ?
what is the function ?
Use parentheses if needed.
as written,
-y - xy' = 3y^2 y'
(3y^2+x)y' = -y
y' = -y/(3y^2+x)
To find dy/dx (the derivative of y with respect to x), you can use implicit differentiation. This involves differentiating both sides of the equation with respect to x while treating y as a function of x.
Let's differentiate both sides of the equation with respect to x.
On the left side, the derivative of 4 with respect to x is 0 because it's a constant.
The derivative of -xy with respect to x can be found using the product rule. The product rule states that the derivative of the product of two functions, u(x) and v(x), is equal to the derivative of u(x) multiplied by v(x), plus the derivative of v(x) multiplied by u(x).
Applying the product rule to -xy, the derivative of -xy with respect to x is (-y)(dx/dx) + (-x)(dy/dx). Simplifying this expression, we have -y + (-x)(dy/dx).
On the right side, we can use the chain rule. The chain rule states that if we have a composite function, such as y^3, the derivative is given by the derivative of the outer function (in this case, y^3) multiplied by the derivative of the inner function (in this case, y).
Applying the chain rule to y^3, the derivative of y^3 with respect to x is 3y^2(dy/dx).
Putting it all together, we have 0 - xy' = 3y^2(dy/dx) - y - x(dy/dx).
Now, we can isolate dy/dx by moving all the terms involving it to one side of the equation:
-xy' + x(dy/dx) = -y + 3y^2(dy/dx).
Next, we can factor out dy/dx from the left side:
y(3y^2 - x) = -y + 3y^2(dy/dx).
Now, we can solve for dy/dx by dividing both sides by (3y^2 - x):
dy/dx = (-y + y(3y^2 - x))/(3y^2 - x).
Simplifying the expression further, we have:
dy/dx = (y(3y^2 - x) - y)/(3y^2 - x).
Finally, the expression for dy/dx in terms of x and y is:
dy/dx = (2y^3 - xy)/(3y^2 - x).