Solve dy/dx for sin(xy)=1+x^3-y^2
sin(xy)=1+x^3-y^2
cos(xy) (y + xy') = 3x^2 - 2yy'
y cos(xy) + xcos(xy) y' = 3x^2 - 2yy'
(x cos(xy) + 2y) y' = 3x^2 - y cos(xy)
y' = (3x^2 - y cos(xy)) / (x cos(xy) + 2y)
To solve the given differential equation, we need to find dy/dx.
Let's start by differentiating both sides of the equation with respect to x using the chain rule.
Differentiating sin(xy) with respect to x gives:
d/dx [sin(xy)] = d/dx [1 + x^3 - y^2]
To calculate the derivative of sin(xy) with respect to x, we break it down using the chain rule:
(cos(xy))(d(xy)/dx) = 3x^2 - 2y(dy/dx)
Using the product rule, we can find d(xy)/dx:
(d(xy)/dx) = (xdy/dx + y)
Now substituting this back into the equation:
(cos(xy))(xdy/dx + y) = 3x^2 - 2y(dy/dx) + 1 + x^3 - y^2
Expanding the equation, we get:
xcos(xy)dy/dx + ycos(xy) = 3x^2 - 2y(dy/dx) + 1 + x^3 - y^2
Rearranging the terms, we can isolate dy/dx:
xcos(xy)dy/dx + 2y(dy/dx) = 3x^2 + x^3 + 1 - y^2 - ycos(xy)
Factoring out dy/dx, we have:
(dy/dx)(xcos(xy) + 2y) = 3x^2 + x^3 + 1 - y^2 - ycos(xy)
Finally, to find dy/dx, we divide both sides by (xcos(xy) + 2y):
dy/dx = (3x^2 + x^3 + 1 - y^2 - ycos(xy)) / (xcos(xy) + 2y)
Thus, the solution to the given differential equation is dy/dx = (3x^2 + x^3 + 1 - y^2 - ycos(xy)) / (xcos(xy) + 2y).