How to Solve a Differential Equation with Nonlinear Terms?

In summary, the given differential equation can be solved by substituting y=vx and then using the product rule and integrating. The final answer is y = xln|lncx|.
  • #1
sara_87
763
0
solve the diffrential equation:
xy' = x(e^(-y/x)) + y


answer:
y = xln|lncx|


problem:
i don't get it

can anyone explain
 
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  • #2
Try re-writing it as

[tex]y^\prime = e^{-y/x} + \frac{y}{x}[/tex]

Does a substitution now spring to mind?
 
  • #3
yeah i get that... but then i don't know what to do... do we substitute y=vx?
 
  • #4
sara_87 said:
yeah i get that... but then i don't know what to do... do we substitute y=vx?
That would indeed be my next step :smile:
 
  • #5
lnC + lnx is lnCx yeah?
 
  • #6
sara_87 said:
lnC + lnx is lnCx yeah?
It is indeed. In general ln(a)+ln(b)=ln(ab).
 
  • #7
ok
no matter what i do I'm not getting that answer! and its not supposed to a hard question!

i get this:

y = x lnCx
 
  • #8
What do you get after the substitution?
 
  • #9
sara_87 said:
ok
no matter what i do I'm not getting that answer! and its not supposed to a hard question!

i get this:

y = x lnCx

What are the last few steps leading up to your answer? Without seeing your work, it is difficult to say where your error is. Though I am inclined to say that you are probably not taking the natural log of the entire right hand side. That is usually where my error is.

Edit: I notice Hootenany has got you covered. You are in good hands. Back to my DE!
 
  • #10
i didn't want to write it out at first because 1. i can't write it the fancy form and 2. I'm affraid i made a very stupid embarrasing mistake, but ok this is what i did:

y' = (e^(-y/x)) + y/x

then y=vx and so v=y/x


d(vx)/dx = (e^(-v)) + v

then i integrate:

vx = x(e^(-v)) + vx + c

then i substituted y back in

is this right so far?
 
  • #11
You need to be careful here since v is not a constant, instead it is a function of x, i.e. v=v(x). Therefore, by the product rule,

[tex]\frac{d}{dx}vx = x\frac{dv}{dx} + v[/tex]

Hence, the second line of your working is actually,

[tex]x\frac{dv}{dx} + v = e^{-v} + v[/tex]

Edit: Just to add a little more detail, since v is a function of x, in general,

[tex]\int v(x)dx \neq v\cdot x + c[/tex]
 
Last edited:
  • #12
yes yes yes! i see

cheers!
 
  • #13
sara_87 said:
yes yes yes! i see

cheers!
No problem :smile:
 

What are differential equations?

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are used to model a wide range of phenomena in various fields such as physics, engineering, economics, and biology.

What is the purpose of solving differential equations?

The main purpose of solving differential equations is to determine the function or functions that satisfy the given equation. This allows us to make predictions and understand the behavior of systems in real-world situations.

What are the different types of differential equations?

There are several types of differential equations, including ordinary differential equations (ODEs), partial differential equations (PDEs), and stochastic differential equations (SDEs). ODEs involve a single independent variable, while PDEs involve multiple independent variables. SDEs take into account random processes and are commonly used in finance and physics.

What are the techniques used to solve differential equations?

There are various techniques used to solve differential equations, including separation of variables, integral transforms, power series, and numerical methods. The choice of technique depends on the type and complexity of the equation.

Why are differential equations important in science?

Differential equations are essential in science because they provide a mathematical framework for understanding and predicting natural phenomena. They are used to model and analyze a wide range of physical systems, from simple pendulums to complex weather patterns.

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