Linearity of ODE: (1 x^2) dy/dx + y = 0

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In summary, the linearity of an ODE is determined by whether or not the sum of any two solutions to it is also a solution. In the given equation, (1-x^2)dy/dx + y = 0, the presence of a square term does not affect the linearity. However, if the coefficient contains the dependent variable, as in (1-y^2), the equation would not be linear.
  • #1
RiceKernel
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Hi , I have no problem to solve but just a bit of confusion on what determines the linearity of an ODE.

Let's say the equation is (1 x^2) dy/dx + y = 0

Is it linear ? I would incline to say yes because the dependent variable and its derivatives are not in a product with each other but the square on the x makes me doubt the linearity or does it not matter at all? If it was (1-y^2), it wouldn't be linear because the coefficient has the dependent variable in it.

Thanks ,
GT
 
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  • #2
An ODE is linear if the sum of any two solutions to it is also a solution.

Suppose we have two functions, ##y_{1}## and ##y_{2}##, and they both satisfy the ODE:

##(1-x^{2})\frac{dy_{1}}{dx} + y_{1}=0##
##(1-x^{2})\frac{dy_{2}}{dx} + y_{2}=0##

Can you show that ##y_{1}+y_{2}## also satisfies the ODE? If you can, then the eq is linear.
 
  • #3
Not helpful at all . Just had to say yes or no .
 
  • #4
give a man a fish...
 
  • #5
RiceKernel said:
Not helpful at all . Just had to say yes or no .
Do the work and you'll get your answer. This isn't a forum for babies, please read the rules.
 

1. What is the definition of linearity in ODEs?

Linearity in ODEs refers to the property of a differential equation where the dependent variable and its derivatives appear in a linear form, meaning they are raised to the first power and are not multiplied or divided by each other. In other words, the equation can be written as a sum of terms, each containing a constant coefficient multiplied by the dependent variable or its derivatives.

2. How do you determine if an ODE is linear or not?

An ODE is considered linear if it can be written in the form of y' + p(x)y = g(x), where p(x) and g(x) are functions of x. This means that the dependent variable and its derivatives appear to the first power and are not multiplied or divided by each other. If the equation cannot be written in this form, it is not considered linear.

3. What is the solution to the ODE (1 x^2) dy/dx + y = 0?

The solution to this ODE is y = C/x, where C is an arbitrary constant. This can be found by solving for y and integrating both sides of the equation. The resulting solution is a family of curves that satisfy the given ODE.

4. Can a linear ODE have nonlinear solutions?

No, a linear ODE can only have linear solutions. This is because the linearity of the ODE is determined by the form of the equation itself, not the specific solution. If a non-linear solution is obtained, it means that the equation is not truly linear.

5. Why is linearity important in ODEs?

Linearity is important in ODEs because it allows us to solve the equation using well-known techniques such as separation of variables, integrating factors, and variation of parameters. It also allows us to use superposition to find the general solution by combining individual solutions. In addition, many real-world applications can be modeled and solved using linear ODEs.

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