Solving Exact Differential Equations: An Attempt at a Solution

In summary, the conversation discusses solving a differential equation by using a substitution method and then integrating the resulting function. The exact equation method is considered, but a substitution of y=vx is used instead. The conversation also covers how to solve for the constant of integration by using natural logs.
  • #1
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Homework Statement



[tex]\frac{dy}{dx}=\frac{x+3y}{3x+y}[/tex]

Homework Equations



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

Exact equation (I think)

The Attempt at a Solution



Ok, so I try to put it in the form of the of an exact equation, like above, but I end up with

-(x+3y)dx+(3x+y)dy=0

This results in the partial of M with respect to y being -3 and the Partial of N with respect to x being 3, which is not equal!

I tried investigating to see if there is an integrating factor but [tex]\frac{M_{y}-N_{x}}{N}[/tex] and [tex]\frac{N_{x}-M_{y}}{M}[/tex] are not functions of just one variable. What am I doing wrong? Different method? Algebra mistake?

Thanks
 
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  • #2
Try using a substitution of y=vx.
 
  • #3
ok, so

[tex]y=vx[/tex]

and

[tex]dy=vdx+xdv[/tex]



[tex](-x-3vx)dx+(3x+vx)(vdx+xdv)=0[/tex] So far so good...



[tex]-xdx+xv^{2}dx+3x^{2}dv+vx^{2}dv=0[/tex] Collect like terms...



[tex]x(v^{2}-1)dx+x^{2}(3+v)dv=0[/tex] Factor out like terms...



[tex]\frac{1}{x}dx=\frac{-(3+v)}{(v^{2}-1)}dv[/tex] Seperate variables...is this right? I tend to make a lot of simple errors, no matter how many times I check my work...

Thanks!
 
Last edited:
  • #4
That should be correct. Now just split the function in 'v' into partial fractions and integrate.
 
  • #5
Ok so...

[tex]\frac{1}{x}=\frac{1}{v+1}-\frac{2}{v-1}[/tex]

[tex]\int\frac{1}{x}dx=\int(\frac{1}{v+1}-\frac{2}{v-1})dv[/tex]

[tex]ln(x)=ln(v+1)-2ln(v-1)+ln(c)[/tex] Ok I don't really know what to do next, here's a try...

[tex]ln(x)=ln(c\frac{v+1}{(v-1)^{2}})[/tex]

Is this right? I'm not very good with natural logs, hopefully I didn't make a mistake...

Next step would be exponentiate both sides? (Assuming what I just did was right)
 
  • #6
Well you don't really have to exponentiate both sides since if lnx=lny then x=y.

So essentially, you can just remove the 'ln' and then replace 'v'.
 

What is a differential equation?

A differential equation is a mathematical equation that describes how one or more variables change over time, based on their rates of change in relation to each other. It is typically used to model natural phenomena and physical systems.

What is the difference between an ordinary and partial differential equation?

An ordinary differential equation involves only one independent variable, while a partial differential equation involves multiple independent variables. In other words, an ordinary differential equation describes the relationship between a function and its derivatives with respect to a single variable, while a partial differential equation describes the relationship between a multivariable function and its partial derivatives with respect to each variable.

What are the applications of differential equations?

Differential equations are used in many fields of science, including physics, engineering, biology, economics, and chemistry. They are used to model and understand various phenomena, such as population growth, chemical reactions, motion of objects, and electrical circuits.

What are the methods for solving differential equations?

There are several methods for solving differential equations, including separation of variables, substitution, and using integrating factors. Other techniques such as Laplace transforms, power series, and numerical methods can also be used depending on the type and complexity of the equation.

Are differential equations important in real life?

Yes, differential equations are very important in real life as they allow us to understand and predict how systems change over time. They are used in various fields, including engineering, economics, and biology, to model and analyze real-world problems and make informed decisions.

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