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Getting term repeated extra time

find_the_fun

Active member
Feb 1, 2012
166
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
Feb 13, 2012
1,704
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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