# Getting term repeated extra time

#### find_the_fun

##### Active member
Solve the differential equation $$\displaystyle (5x+4y)dx+(4x-8y^3)dy=0$$

So $$\displaystyle M(x, y) = 5x+4y$$ and $$\displaystyle N(x, y) = 4x-8y^3$$

Check $$\displaystyle \frac{\partial M}{\partial y} = 4$$ and $$\displaystyle \frac{ \partial N}{\partial x} = 4$$ check passed.

$$\displaystyle f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y)$$

$$\displaystyle \frac{\partial f(x, y)}{\partial y} = g'(y) = 4x-8y^3$$

Therefore $$\displaystyle g(y) = 4xy-2y^4$$

So $$\displaystyle f(x, y)=\frac{5x^2}{2}+4yx+4yx-2y^4$$

The back of book has only one 4yx so what did I do wrong? Also the back of the book has the equation $$\displaystyle =C$$ and I don't understand why?

#### chisigma

##### Well-known member
Solve the differential equation $$\displaystyle (5x+4y)dx+(4x-8y^3)dy=0$$

So $$\displaystyle M(x, y) = 5x+4y$$ and $$\displaystyle N(x, y) = 4x-8y^3$$

Check $$\displaystyle \frac{\partial M}{\partial y} = 4$$ and $$\displaystyle \frac{ \partial N}{\partial x} = 4$$ check passed.

$$\displaystyle f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y)$$

$$\displaystyle \frac{\partial f(x, y)}{\partial y} = g'(y) = 4x-8y^3$$

Therefore $$\displaystyle g(y) = 4xy-2y^4$$

So $$\displaystyle f(x, y)=\frac{5x^2}{2}+4yx+4yx-2y^4$$

The back of book has only one 4yx so what did I do wrong? Also the back of the book has the equation $$\displaystyle =C$$ 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

Staff member
Solve the differential equation $$\displaystyle (5x+4y)dx+(4x-8y^3)dy=0$$

So $$\displaystyle M(x, y) = 5x+4y$$ and $$\displaystyle N(x, y) = 4x-8y^3$$

Check $$\displaystyle \frac{\partial M}{\partial y} = 4$$ and $$\displaystyle \frac{ \partial N}{\partial x} = 4$$ check passed.

$$\displaystyle f(x, y) = \int M \partial x + g(y) = \int 5x+4y \partial x + g(y) = \frac{5x^2}{2} + 4yx + g(y)$$
At this point, when you take the partials with respect to $y$, you should get:

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

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

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

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

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