Solve the differential equation (x^4 - 2xy^2 + y^4) dx - (2x^2y - 4xy^3 + sin y) dy = 0.

Steps to Solve

1. Identify the differential equation The given equation is $$(x^4 - 2xy^2 + y^4)dx - (2x^2y - 4xy^3 + \sin y)dy = 0.$$ We can compare it to the standard form ( M dx + N dy = 0 ) where: $$ M = x^4 - 2xy^2 + y^4 $$ $$ N = - (2x^2y - 4xy^3 + \sin y) $$

2. Check for exactness To determine if the differential equation is exact, we need to verify if $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}. $$

Calculate ( \frac{\partial M}{\partial y} ): $$ \frac{\partial M}{\partial y} = -4xy + 4y^3 $$

Now calculate ( \frac{\partial N}{\partial x} ): $$ \frac{\partial N}{\partial x} = - (4xy - 2x^2) $$

You can simplify and compare these results.

3. Integrate M with respect to x If the equation is exact, integrate ( M ) with respect to ( x ): $$ \psi(x, y) = \int (x^4 - 2xy^2 + y^4)dx = \frac{x^5}{5} - x^2y^2 + xy^4 + g(y) $$ where ( g(y) ) is an arbitrary function of ( y ).

4. Differentiate with respect to y Differentiate ( \psi(x, y) ) with respect to ( y ): $$ \frac{\partial \psi}{\partial y} = -2xy + 4xy^3 + g'(y) $$

5. Set equal to N Equate this to ( N ) to find ( g'(y) ): $$ -2xy + 4xy^3 + g'(y) = - (2x^2y - 4xy^3 + \sin y) $$ Solve for ( g'(y) ) and integrate to find ( g(y) ).

6. Combine results The final solution can be written as: $$ \psi(x, y) = C $$ where ( C ) is a constant. Substitute ( g(y) ) and ( \psi(x, y) ) back to form the complete solution.