Solving Complex Differential Equations: What Are the Steps?

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In summary, the conversation is about solving differential equations, specifically x^2y' + cos(2y) = 1 and y' = (4x + 2y - 1)^1/2. The participants discuss using trigonometric identities and separating variables to solve the equations. They also mention one remaining equation that resembles a Riccati equation, but they are unsure how to approach it. Lastly, one participant suggests squaring the equation to start off and using a substitution to solve it.
  • #1
Icebreaker
How do I go about solving

x^2y' - cos (2y) = 1

This is unlike anything I've seen so far. Or so I think.
 
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  • #2
Icebreaker said:
How do I go about solving

x^2y' + cos (2y) = 1

This is unlike anything I've seen so far. Or so I think.

It's separable right?
 
  • #3
saltydog said:
It's separable right?
Yes
Trig is your friend.
as a matter of personal taste I enjoy the use of the identity
1-cos(2y)=2(sin(y/2))^2
 
  • #4
Ok, I got it, thanks for the help.

I got y = arctan ( c = 2/x)
 
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  • #5
18 down, 2 more to go:

y' = (4x + 2y - 1)^(1/2)

and

(1-xy)y' + y ^2 + 3xy^3 = 0

The last one kinda looks like Riccati, but I can't get it to work.


edit: I got the second one. One more ODE remains... I don't know how to approach that one.
 
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  • #6
Icebreaker said:
18 down, 2 more to go:

y' = (4x + 2y - 1)^(1/2)

and

(1-xy)y' + y ^2 + 3xy^3 = 0

The last one kinda looks like Riccati, but I can't get it to work.


edit: I got the second one. One more ODE remains... I don't know how to approach that one.

Regarding:

[tex]\frac{dy}{dx}=\sqrt{4x+2y-1}[/tex]

How about squaring it for starters. Then let:

[tex]p=\frac{dy}{dx}[/tex]

Differentiate it to get an ODE in p. Then separate variables.
 

What is the equation "Dif Eq x^2y' - cos(2y) = 1" used for?

The equation "Dif Eq x^2y' - cos(2y) = 1" is used to model and solve problems involving exponential growth or decay, such as population growth, radioactive decay, and economic growth.

What is the meaning of the variables in "Dif Eq x^2y' - cos(2y) = 1"?

In this equation, x represents the independent variable, y represents the dependent variable, and y' represents the derivative of y with respect to x.

What is the general solution to "Dif Eq x^2y' - cos(2y) = 1"?

The general solution to this equation is y(x) = ±√(C + x^2 + sin(2y)), where C is a constant of integration.

Can "Dif Eq x^2y' - cos(2y) = 1" be solved analytically?

Yes, this equation can be solved analytically using various methods such as separation of variables, integrating factor, or substitution.

What are some real-life applications of "Dif Eq x^2y' - cos(2y) = 1"?

This equation can be applied to study and predict growth or decay in various fields such as biology, chemistry, physics, and economics. It can also be used to model oscillations and periodic phenomena.

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