Find the general solution of this nonlinear ODE: xy' + sin(2y) = x^3*sin^2(y)

In summary, this equation is non linear, non separable, and weird. I would like to have a direction to start working on this.
  • #1
Kaguro
221
57
Homework Statement
Find the general solution of:

xy' + sin(2y) = x^3*sin^2(y)
Relevant Equations
All math.
This equation, is non linear, non-separable, and weird. I would like to have a direction to start working on this.
I tried writing sin(2y) = 2sin(y)*cos(y).
See,
##xy' = x^3sin^2(y)-2sin(y)cos(y)##

Can't separate.

Writing in this way:
##(x^3sin^2y-sin2y)dx-xdy=0##

Also, I checked that it is not exact.
So, what next should I try?
 
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  • #2
How about ##\sin(2y)=2\sin(y)\cos(y)=(\sin^2y)'##?
 
  • #3
fresh_42 said:
How about ##\sin(2y)=2\sin(y)\cos(y)=(\sin^2y)'##?

No. I don't see how that helps...

##x\frac{dy}{dx}+\frac{d(sin^2y)}{dy}=x^3sin^2y##

I can hardly take the LCM of dy and dx...
 
  • #4
Kaguro said:
No. I don't see how that helps...

##x\frac{dy}{dx}+\frac{d(sin^2y)}{dy}=x^3sin^2y##

I can hardly take the LCM of dy and dx...
Yes, it's quite tricky. I looked up the solution and the only other idea that gave me is to try a Weierstrass substitution:
https://en.wikipedia.org/wiki/Weierstrass_substitution
 
  • #5
Your d.e. is non-linear and not amenable to analytic integration. I think you have two options:
a.) Solve it by numerical integration which could be quite tricky considering the non-linearities involved.
b.) Assume there was a misprint in the problem statement (it happens), and solve the equation:$$
xy' + \sin(2x) = x^3\sin^2(x)$$
Option a.) makes sense if you are in a numerical methods class. Option b.) makes sense if you are in an introductory class for differential equations and the instructor has sloppy handwriting.
 
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  • #6
Fred Wright said:
Your d.e. is non-linear and not amenable to analytic integration. I think you have two options:
a.) Solve it by numerical integration which could be quite tricky considering the non-linearities involved.
b.) Assume there was a misprint in the problem statement (it happens), and solve the equation:$$
xy' + \sin(2x) = x^3\sin^2(x)$$
Option a.) makes sense if you are in a numerical methods class. Option b.) makes sense if you are in an introductory class for differential equations and the instructor has sloppy handwriting.
WolframAlpha had a closed analytic solution. But I don't have the pro version to look up the steps. However, it looks as if Weierstraß could help. Otherwise there is still the option to simply proof that the solution is one by differentiation.
 
  • #7
The question was from a daily assignment for a competitive exam. I'm not supposed to do it numerically. But it has a very high probability that there was some typing mistake...(since I'm supposed to be able to solve this question under 4 minutes..)

Also, I've never used Weierstrass substitution...
 
  • #8
Kaguro said:
(since I'm supposed to be able to solve this question under 4 minutes..)
Do you know the derivative of ##\operatorname{arccot}## by heart?
 
  • #9
Yes. I keep getting such integrals to solve, so I need to memorize some of these basic tables by heart.
 
  • #10
The solution is ##-\operatorname{arccot}p(x)## with a polynomial in ##x##. So differentiate and see what you can find out about ##p##.
 
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  • #11
Given, y= -arccot(p)
##y' = \frac{p'}{1+p^2}##
##sin(2x) = 2sinxcosx=\frac{2sinx/cosx}{1/cos^2 (x)} =\frac{2tanx}{sec^2 (x)}= \frac{2tanx}{1+tan^2 (x)}##
sin(-2arccot(p)) =-2p/(1+p^2)

similarly, sin^2(x) = tan^2(x)/(1+tan^2(x))
sin^2(-arccotp) = 1/(1+p^2)
So, finally:
##\frac{xp'}{1+p^2} - \frac{2p}{1+p^2} = \frac{x^3}{1+p^2}##
##xp' - 2p = x^3##
##p' - 2p/x = x^2##

Which I solved by integrating factor as 1/x^2.
p = x^3 + Cx^2

Is this ok?
And how on Earth did you find the form of the solution?
 
  • #12
Looks good.
I used an internet tool to look up the solution:
https://www.wolframalpha.com/input/?i=xy'=x^3+sin^2(y)+-+2+sin(y)+cos(y)

But that was why I thought the Weierstraß- or half-tangent substitution could help. It is generally a good idea if terms in ##x## and trigonometric functions in ##x## occur, because it turns trig functions into polynomials:
https://de.wikipedia.org/wiki/Weierstraß-Substitution#Beschreibung_der_Substitution
is easier to read than the English version. You only need to look at the formulas.
 
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  • #13
If you divide the original differential equation by ##\sin^2 y##, you get
$$x y' \csc^2 y = x^3 - \frac{2 \sin y \cos y}{\sin^2 y}.$$ Note that ##y' \csc^2 y = -(\cot y)'##. Now you have a first-order differential equation in ##\cot y##.
 
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  • #14
vela said:
If you divide the original differential equation by ##\sin^2 y##, you get
$$x y' \csc^2 y = x^3 - \frac{2 \sin y \cos y}{\sin^2 y}.$$ Note that ##y' \csc^2 y = -(\cot y)'##. Now you have a first-order differential equation in ##\cot y##.
Oh! So that's how one gets the idea to use a trial solution of arccot!

That's great!

You guys are so powerful, manipulating equation in ways unimaginable by me...
 

FAQ: Find the general solution of this nonlinear ODE: xy' + sin(2y) = x^3*sin^2(y)

1. What is a nonlinear ODE?

A nonlinear ODE (ordinary differential equation) is a mathematical equation that involves derivatives of a function with respect to its independent variable, and the function itself is not a linear combination of its derivatives.

2. How do you find the general solution of a nonlinear ODE?

To find the general solution of a nonlinear ODE, you can use various methods such as separation of variables, substitution, or integrating factors. However, for more complex nonlinear ODEs, it may not be possible to find an explicit solution and numerical methods may be needed.

3. What is the general solution of the given nonlinear ODE?

The general solution of the given nonlinear ODE, xy' + sin(2y) = x^3*sin^2(y), is y = -cos^-1(C/x) or y = pi/2 + cos^-1(C/x), where C is a constant of integration.

4. How do you verify the general solution of a nonlinear ODE?

To verify the general solution of a nonlinear ODE, you can substitute the solution into the original equation and see if it satisfies the equation. In the case of the given ODE, substituting y = -cos^-1(C/x) or y = pi/2 + cos^-1(C/x) into xy' + sin(2y) = x^3*sin^2(y) should result in a true statement.

5. Can a nonlinear ODE have multiple solutions?

Yes, a nonlinear ODE can have multiple solutions. This is because the general solution of a nonlinear ODE typically includes a constant of integration, which can take on different values and result in different solutions. In the case of the given ODE, there are two possible solutions depending on the value of the constant C.

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