Getting term repeated extra time

[find_the_fun]

Solve the differential equation [math](5x+4y)dx+(4x-8y^3)dy=0[/math]

So [math]M(x, y) = 5x+4y[/math] and [math]N(x, y) = 4x-8y^3[/math]

Check [math]\frac{\partial M}{\partial y} = 4[/math] and [math]\frac{ \partial N}{\partial x} = 4 [/math] check passed.

[math]f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y) [/math]

[math]\frac{\partial f(x, y)}{\partial y} = g'(y) = 4x-8y^3[/math]

Therefore [math]g(y) = 4xy-2y^4[/math]

So [math]f(x, y)=\frac{5x^2}{2}+4yx+4yx-2y^4[/math]

The back of book has only one 4yx so what did I do wrong? Also the back of the book has the equation [math]=C[/math] and I don't understand why?

 

[chisigma]

Solve the differential equation [math](5x+4y)dx+(4x-8y^3)dy=0[/math]

So [math]M(x, y) = 5x+4y[/math] and [math]N(x, y) = 4x-8y^3[/math]

Check [math]\frac{\partial M}{\partial y} = 4[/math] and [math]\frac{ \partial N}{\partial x} = 4 [/math] check passed.

[math]f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y) [/math]

[math]\frac{\partial f(x, y)}{\partial y} = g'(y) = 4x-8y^3[/math]

Therefore [math]g(y) = 4xy-2y^4[/math]

So [math]f(x, y)=\frac{5x^2}{2}+4yx+4yx-2y^4[/math]

The back of book has only one 4yx so what did I do wrong? Also the back of the book has the equation [math]=C[/math] and I don't understand why?

The solution is of the form...

$\displaystyle \int M\ dx + \int (N - \frac{d}{dy} \int M\ dx)\ dy = c\ (1)$

... and with little effort You find...

$\displaystyle \frac{5}{2}\ x^{2} + 4\ x\ y - 2\ y^{4} = c\ (2)$

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

Solve the differential equation [math](5x+4y)dx+(4x-8y^3)dy=0[/math]

So [math]M(x, y) = 5x+4y[/math] and [math]N(x, y) = 4x-8y^3[/math]

Check [math]\frac{\partial M}{\partial y} = 4[/math] and [math]\frac{ \partial N}{\partial x} = 4 [/math] check passed.

[math]f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y) [/math]

At this point, when you take the partials with respect to $y$, you should get:

$$4x-8y^3=4x+g'(y)$$

$$g'(y)=-8y^3$$

$$g(y)=-2y^4$$

And now, you take the solution as given implicitly by:

$$F(x,y)=C$$