Solving 1st Order ODE with Homogeneous Techniques | Initial Value y(0)=0

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In summary, the conversation discusses solving the equation dy/dx = x^2 + y^2 with the initial condition y(0) = 0 using the method of substitution and integrating to find the solution. The constant term can be found by plugging in the initial condition, and there may be discrepancies in the Picard's iteration method due to rounding errors or mistakes in the process.
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dglee
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dy/dx=x^2+y^2; y(0)=0

okay i solved this question using homogenous

dy/dx=1+(y/x)^2

subsitute v=y/x

v+x*dv/dx=1+v^2 then use exact equation

dv/(v^2-v+1)=dx/x

dv/((v-1/2)^2+3/4)=(dx/x or just ln(x))

then integrate dv/((v-1/2)^2+3/4)

u=v-1/2
du=dv

dv/(u^2)+3/4

s=2u/sqrt(3)
ds=2/sqrt(3)du

thus

2/sqrt[3]*arctan=ln[x]
then substitue everything back in

2/sqrt[3]*arctan[2u/sqrt[3]]=ln[x]+c

2/sqrt[3]*arctan[2(v-1/2)/sqrt[3]]=ln[x]+c

2/sqrt[3]*arctan[2(y/x-1/2)/sqrt[3]]=ln[x]+c

okay ths thing is now i can not plug in y(0)=0 to get the constant for c and the other thing is this is suppose to be a unique and exisit throughout the interval [0,1/2] i used picard's theorem to show that its continus and estimated h=[0,1/2] the thing is I am suppose to plug in 1/2 into y
y(1/2) and when i solve the equation however i can't get a constant
and when i use picard's iteration method successful approximation i do not get the same number. well one thing is i don't have the constant c unless i did the ODE totally wrong. can't think of another first order technique.

when i do picard's iteration method i get .047**** but if i have c i could might have the same answer when i plug in 1/2 one question is am i suppose to do solving it by subtitution and how is it possible tof ind the constant term? or am i totally wrong?
 
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Hello! Thank you for sharing your solution to this problem. It looks like you have approached it correctly using the method of substitution and integrating to find the solution.

As for finding the constant term, you can use the initial condition given in the problem to solve for it. In this case, we know that y(0) = 0, so we can plug that into the equation to get:

2/sqrt[3]*arctan[2(0-1/2)/sqrt[3]]=ln[0]+c

-2/sqrt[3]*arctan[1/sqrt[3]]=c

c = -2/sqrt[3]*arctan[1/sqrt[3]]

So, your final solution would be:

2/sqrt[3]*arctan[2(y/x-1/2)/sqrt[3]]=ln[x]-2/sqrt[3]*arctan[1/sqrt[3]]

As for the discrepancy in your Picard's iteration method, it could be due to rounding errors or a mistake in the iteration process. I would suggest double checking your calculations and making sure you are using the correct initial value for your first iteration.

Overall, it seems that you have a good understanding of how to approach this problem and have used the correct techniques. Keep up the good work!
 

1. What is a 1st order ODE?

A 1st order ODE (ordinary differential equation) is a type of mathematical equation that involves one independent variable and its derivatives. It is usually in the form of y' = f(x,y) where y' is the derivative of y with respect to x.

2. What does it mean for a 1st order ODE to be homogeneous?

A 1st order ODE is considered homogeneous if all of its terms have the same degree. In other words, each term in the equation has the same power of the dependent variable and its derivatives.

3. What are the techniques used to solve 1st order ODEs with homogeneous coefficients?

The most common techniques used to solve 1st order ODEs with homogeneous coefficients are separation of variables, substitution, and the use of integrating factors. These techniques involve manipulating the equation to separate the variables and integrating to find the solution.

4. What does the initial value y(0)=0 mean in the context of solving a 1st order ODE?

The initial value y(0)=0 means that the value of the dependent variable y at the initial point x=0 is equal to 0. This is a specific condition that is given in the problem and is used to find the particular solution to the 1st order ODE.

5. Why is it important to specify an initial value when solving a 1st order ODE?

Specifying an initial value is important because it allows us to find the particular solution to the 1st order ODE. Without an initial value, the solution would have an arbitrary constant and would not be unique. The initial value helps to determine the exact value of the constant and find the specific solution for the given problem.

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