Solving Exact Diff. Eq: (5x+4y)dx+(4x-8y^3)dy=0

[find_the_fun]

Determine whether the given differential equation is exact. If it is, solve it.

\(\displaystyle (5x+4y)dx+(4x-8y^3)dy=0\)

So \(\displaystyle M(x, y) = 5x+4y\) and \(\displaystyle N(x, y)=4x-8y^3\)
Then \(\displaystyle \frac{\partial M}{\partial Y} = 4\) and \(\displaystyle \frac{\partial N}{\partial y} = 4\) therefore the equation is exact and there exists a function f such that

\(\displaystyle \frac{\partial f}{\partial x} = 5x+4y\) and \(\displaystyle \frac{ \partial f}{\partial y} = 4x-8y^3\)

\(\displaystyle \int 5x+4 dx = \frac{5}{2}x^2+4xy=g(y) = f(x,y)\)
\(\displaystyle \frac{\partial f}{\partial y} = 4x+g'(y) = 4x-8y^3\)

Now how do I solve for g(y)?

 

[Euge]

Determine whether the given differential equation is exact. If it is, solve it.

\(\displaystyle (2x+y)dx-(x+6y)dy=0\)

So \(\displaystyle M(x, y) = 5x+4y\) and \(\displaystyle N(x, y)=4x-8y^3)\)
Then \(\displaystyle \frac{\partial M}{\partial Y} = 4\) and \(\displaystyle \frac{\partial N}{\partial y} = 4\) therefore the equation is exact and there exists a function f such that

\(\displaystyle \frac{\partial f}{\partial x} = 5x+4y\) and \(\displaystyle \frac{ \partial f}{\partial y} = 4x-8y^3\)

\(\displaystyle \int 5x+4 dx = \frac{5}{2}x^2+4xy=g(y) = f(x,y)\)
\(\displaystyle \frac{\partial f}{\partial y} = 4x+g'(y) = 4x-8y^3\)

Now how do I solve for g(y)?

How did you get $N(x,y) = 4x - 8y^3$? It should be $N(x,y) = -(x + 6y)$, in which case $N_x = -1 \neq 1 = M_y$ and hence the equation is inexact.

Are you showing two different problems, one which is $(2x + y) dx - (x + 6y) dy = 0$ and the other which is $(5x + 4y) dx + (4x - 8y^3) dy = 0$?

 

[find_the_fun]

How did you get $N(x,y) = 4x - 8y^3$? It should be $N(x,y) = -(x + 6y)$, in which case $N_x = -1 \neq 1 = M_y$ and hence the equation is inexact.

Are you showing two different problems, one which is $(2x + y) dx - (x + 6y) dy = 0$ and the other which is $(5x + 4y) dx + (4x - 8y^3) dy = 0$?

Sorry I copied the wrong question. It's fixed now.

 

[Euge]

Determine whether the given differential equation is exact. If it is, solve it.

\(\displaystyle (5x+4y)dx+(4x-8y^3)dy=0\)

So \(\displaystyle M(x, y) = 5x+4y\) and \(\displaystyle N(x, y)=4x-8y^3\)
Then \(\displaystyle \frac{\partial M}{\partial Y} = 4\) and \(\displaystyle \frac{\partial N}{\partial y} = 4\) therefore the equation is exact and there exists a function f such that

\(\displaystyle \frac{\partial f}{\partial x} = 5x+4y\) and \(\displaystyle \frac{ \partial f}{\partial y} = 4x-8y^3\)

\(\displaystyle \int 5x+4 dx = \frac{5}{2}x^2+4xy=g(y) = f(x,y)\)
\(\displaystyle \frac{\partial f}{\partial y} = 4x+g'(y) = 4x-8y^3\)

Now how do I solve for g(y)?

From $4x + g'(y) = 4x - 8y^3$, we get $g'(y) = -8y^3$. By integration, $g(y) = -2y^4 + C$, where $C$ is an arbitrary constant.

 

[find_the_fun]

From $4x + g'(y) = 4x - 8y^3$, we get $g'(y) = -8y^3$. By integration, $g(y) = -2y^4 + C$, where $C$ is an arbitrary constant.

Ok. My textbook does a really poor job at explaining this step but I notice it doesn't include $C$. Can it be omitted for some reason, for example because you set the entire implicit equation equal to an arbitrary constant?

For example the answer is \(\displaystyle \frac{5}{2}x^2+4xy-2y^4=c\) and not\(\displaystyle \frac{5}{2}x^2+4xy-2y^4+c_1=c_2\)

 

[Euge]

Ok. My textbook does a really poor job at explaining this step but I notice it doesn't include $C$. Can it be omitted for some reason, for example because you set the entire implicit equation equal to an arbitrary constant?

Yes, that's right.

 

[find_the_fun]

Yes, that's right.

Actually I don't quite get what's going on. It seems like with exact equations we go through the steps to find a function f(x, y) we pulled out of thin air and now we say it's equal to some constant. How do we know it's equal to a constant? That basically means it doesn't change in the z-direction right?

 

[find_the_fun]

Also the answer key often multiplies through by -1 but c is still c and not -c, is that the same reason?

 

[mathbalarka]

$C$ is an arbitrary real constant, right? Then $-C$ is also an arbitrary real constant. You can replace it by $C$ without loss of generality.

 

[Euge]

Actually I don't quite get what's going on. It seems like with exact equations we go through the steps to find a function f(x, y) we pulled out of thin air and now we say it's equal to some constant. How do we know it's equal to a constant? That basically means it doesn't change in the z-direction right?

Suppose you have an exact equation $M(x,y)\, dx + N(x,y)\,dy = 0$ in a domain $D \subset \Bbb R^2$. There is an $f : D \to \Bbb R$ such that $f_x = M$ and $f_y = N$. So the differential $df = f_x\, dx + f_y\, dy = M\, dx + N\, dy = 0$. The solutions to $df = 0$ are the level curves of the surface $z = f(x,y)$ in $D \times \Bbb R$. For every constant $C$, $d(f + C) = df + dC = df + 0 = 0$. It follows that for every constant $C$, the set level curves of $z = f(x,y)$ is equal to the set of level curves of $z = f(x,y) + C$. So the one-parameter family of solutions $f(x,y) = \text{constant}$ is unaffected by a translation in the $z$-direction.