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

In summary: But, using the chain rule, if F(x,y) is a constant, then\frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}= 0for any t. That can only be true for all t if both\frac{\partial F}{\partial x}= \frac{dy}{dt}= 0and\frac{\partial F}{\partial y}= \frac{dy}{dt}= 0 Now, consider the equation we are trying to solve in differential form:M(x,y)dx+ N(x,y)dy= 0
  • #1
hover
343
0

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!
 
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  • #2
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'.
 
  • #3
hover said:

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
[tex]\frac{y+ 4y^4}{y^4}dx- \frac{y^3+ 3x}{y^4}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy[/tex]
is an exact differential. That means that there must exist some function F(x, y) such that
[tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy[/tex]

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

The first is easy. Integrating with respect to x (treating y as a constant)
[tex]F(x,y)= (y^{-3}+ 4)x+ f(y)= xy^{-3}+ 4x+ f(y)[/tex]
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:
[tex]\frac{\partial F}{\partial y}= -3xy^{-4}+ f'(y)[/tex]
and that must be equal to
[tex]-y^{-1}- 3xy^{-4}[/tex]

Set them equal, solve for f' and then f.
 
  • #4
HallsofIvy said:
No, you cannot just combine two functions like that. You are saying that
[tex]\frac{y+ 4y^4}{y^4}dx- \frac{y^3+ 3x}{y^4}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy[/tex]
is an exact differential. That means that there must exist some function F(x, y) such that
[tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy= (y^{-3}+ 4)dy- (y^{-1}- 3xy^{-4})dy[/tex]

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

The first is easy. Integrating with respect to x (treating y as a constant)
[tex]F(x,y)= (y^{-3}+ 4)x+ f(y)= xy^{-3}+ 4x+ f(y)[/tex]
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:
[tex]\frac{\partial F}{\partial y}= -3xy^{-4}+ f'(y)[/tex]
and that must be equal to
[tex]-y^{-1}- 3xy^{-4}[/tex]

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.
 
  • #5
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
[tex]\frac{dF}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}[/tex]

In "differential form",
[tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy[/tex]

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

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It involves the use of derivatives to represent the rate of change of a quantity over time or space.

2. Why are differential equations important?

Differential equations are important because they are used to model and understand many natural phenomena in science and engineering, such as motion, growth, and decay. They also play a crucial role in many areas of modern technology, including control systems, signal processing, and computer graphics.

3. How do you solve a differential equation?

The method for solving a differential equation depends on its type and order. Some common techniques include separation of variables, substitution, and using integrating factors. In some cases, a numerical approach may be necessary. It is also important to check for any initial or boundary conditions when solving a differential equation.

4. What is the difference between ordinary and partial differential equations?

An ordinary differential equation (ODE) involves a single independent variable and its derivatives, while a partial differential equation (PDE) involves multiple independent variables and their derivatives. ODEs are used to model phenomena in a single dimension, while PDEs are used for multi-dimensional phenomena.

5. Where are differential equations used in real life?

Differential equations are used in many fields, including physics, chemistry, biology, economics, and engineering. They are used to model and predict the behavior of systems such as population growth, heat transfer, fluid dynamics, and electric circuits. They are also used in the development of computer models and simulations.

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