How can the chain rule be applied to solve this differential equation?

[hover]

Homework Statement

The differential equation y + 4y^4 = (y^3 + 3x)y' can be written in differential form:

M(x,y)dx + N(x,y)dy = 0

where

M(x,y)= y+4y^4 , and N(x,y)= -y^3-3x

The term (M(x,y)dx + N(x,y) dy) becomes an exact differential if the left hand side above is divided by y^4. Integrating that new equation, the solution of the differential equation is "answer goes here" = C

Homework Equations

The Attempt at a Solution

Am I suppose to do this and integrate?

(y+4y^4-y^3-3x)/y^4 = 0

or am I off the ball completely?

Thanks!

 

[rock.freak667]

I do not think y⁴ is an integrating factor but I did not check if ∂M/∂y=∂N/∂x

But if f(x,y) is the solution to the exact DE, then

∂f/∂x=M and ∂f/∂y=N

So you can find f(x,y) from either and then you use the other one to get the function in either y or x.

e.g. ∂f/∂x=3x², then f(x,y)=x³+h(y), in which I would then put ∂f/∂y=N=h'(y) and integrate and solve for 'y'.

 

[HallsofIvy]

Homework Statement

The differential equation y + 4y^4 = (y^3 + 3x)y' can be written in differential form:

M(x,y)dx + N(x,y)dy = 0

where

M(x,y)= y+4y^4 , and N(x,y)= -y^3-3x

The term (M(x,y)dx + N(x,y) dy) becomes an exact differential if the left hand side above is divided by y^4. Integrating that new equation, the solution of the differential equation is "answer goes here" = C

Homework Equations

The Attempt at a Solution

Am I suppose to do this and integrate?

(y+4y^4-y^3-3x)/y^4 = 0

or am I off the ball completely?

Thanks!

No, you cannot just combine two functions like that. You are saying that
$$\frac{y+ 4y^4}{y^4}dx- \frac{y^3+ 3x}{y^4}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy$$
is an exact differential. That means that there must exist some function F(x, y) such that
$$dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy$$

That means you need to solve
$$\frac{\partial F}{\partial x}= \frac{y+ 4y^4}{y^4}= y^{-3}+ 4$$
and
$$\frac{\partial F}{\partial y}= \frac{-y^3- 3x}{y^4}= -y^{-1}- 3xy^{-4}$$

The first is easy. Integrating with respect to x (treating y as a constant)
$$F(x,y)= (y^{-3}+ 4)x+ f(y)= xy^{-3}+ 4x+ f(y)$$
where f can be any function of y (the derivative of any function of y with respect to x is 0).

Now differentiate that with respect to y:
$$\frac{\partial F}{\partial y}= -3xy^{-4}+ f'(y)$$
and that must be equal to
$$-y^{-1}- 3xy^{-4}$$

Set them equal, solve for f' and then f.

 

[hover]

No, you cannot just combine two functions like that. You are saying that
$$\frac{y+ 4y^4}{y^4}dx- \frac{y^3+ 3x}{y^4}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy$$
is an exact differential. That means that there must exist some function F(x, y) such that
$$dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy$$

That means you need to solve
$$\frac{\partial F}{\partial x}= \frac{y+ 4y^4}{y^4}= y^{-3}+ 4$$
and
$$\frac{\partial F}{\partial y}= \frac{-y^3- 3x}{y^4}= -y^{-1}- 3xy^{-4}$$

The first is easy. Integrating with respect to x (treating y as a constant)
$$F(x,y)= (y^{-3}+ 4)x+ f(y)= xy^{-3}+ 4x+ f(y)$$
where f can be any function of y (the derivative of any function of y with respect to x is 0).

Now differentiate that with respect to y:
$$\frac{\partial F}{\partial y}= -3xy^{-4}+ f'(y)$$
and that must be equal to
$$-y^{-1}- 3xy^{-4}$$

Set them equal, solve for f' and then f.

f'(y) = -y^-1 , f(y) = -ln(y) so F(x,y) = xy^-3+4x-ln(y)

While I was able to figure out what the problem was asking thanks to your help, I feel like I still don't truly understand the question it is asking. Like where did those partial derivatives come from?.. I don't know maybe its just me.

 

[HallsofIvy]

One thing I am sure you learned in Calculus is the "chain rule" for functions of more than one variable. If F is a function of x and y, and x and y are both functions of t, then
$$\frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}$$

In "differential form",
$$dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy$$

Saying that dF= 0 means that dF/dt= 0 for any parameter t and so F is a constant.