Solving separable equation sin3x dx + 2y(cos3x)³ dy = 0

[bobmerhebi]

1. Solve by Seperable Variable: sin3xdx + 2ycos³3xdy = 0

2. We need write it in the form: dy/dx = g(x).h(y)

3. The Attempt at a Solution :

a few algebraic operations lead to: dy/dx = (-sin3x/cos³3x).(1/2y) with h(y) = 1/2y
now we have (y²)' = 2ydy = -sin3xdx/cos³3x
integrating we get: y² = $$\int$$-sin3xdx/cos³3x =G(x)

thus y² = (1/3).$$\int$$d(cos3x)/cos³3x = (1/3).(-1/2).(1/cos²3x)
so y² = -1/6cos²3x.

here i reached a dead end. is there anything wrong in the attempt or is it right as i think & the given d.e. is wrong ? note that i tried to solve it more then 10 times, i also tried the tan3x & its derivative but the -ve sign persists.

waiting for your help. thanks in advance

 

[gabbagabbahey]

Assuming that you mean : $$y^2=\frac{-1}{6cos^2(3x)}$$ , then your solution is correct.

When you differentiate it you get:

$$2ydy=(-2)(3) \left( \frac{sin^3(3x)}{6cos^3(3x)} \right)=\frac{-sin^3(3x)}{cos^3(3x)}$$

which gives you back your original ODE...why were you thinking it was incorrect?

 

[HallsofIvy]

That is a solution to the differential equation. It is not the general solution.

What happened to the constant of integration in $y^2 = -\int sin3xdx/cos33x$

 

[bobmerhebi]

nop but my doctor gave it besides others as ana assignment & i tried it many times - at first yeh i though it was wrong - so i got the same answer over & over again.
So there's some problem with the D.E. given. I use the book: A first Course in Differential equations by Dennis Zill 8th ed.

 

[bobmerhebi]

ofcorse its a solution. there's many sol.s , but eventhough the answer y2 is -ve which is impossible.

 

[tiny-tim]

Welcome to PF!

Hi bobmerhebi! Welcome to PF!

2. We need write it in the form: dy/dx = g(x).h(y)
…
a few algebraic operations lead to: dy/dx = (-sin3x/cos³3x).(1/2y) with h(y) = 1/2y
now we have (y²)' = 2ydy = -sin3xdx/cos³3x

hmm … that took a long time …

it's quicker to start by separating them into the form h(y)dy = g(x)dx …

the clue's in the word "separable"

… thus y² = (1/3).$$\int$$d(cos3x)/cos³3x = (1/3).(-1/2).(1/cos²3x)
so y² = -1/6cos²3x.
…
the -ve sign persists.

Yes, you're wondering how the LHS can be positive and the RHS negative …

that'll teach you not to leave out the constant of integration in future, won't it? ​

 

[bobmerhebi]

so including the constant of integration c we have:

y² = (1/3).( <2cos²3x -1>/2cos²3x)
= (2c.cos²3x - 1)/6cos²3x

where the numerator = 2c-1 but c should be positive right for the solution to be valid?

 

[tiny-tim]

so including the constant of integration c we have:

y² = (1/3).( <2cos²3x -1>/2cos²3x)
= (2c.cos²3x - 1)/6cos²3x

where the numerator = 2c-1 but c should be positive right for the solution to be valid?

Sorry, bobmerhebi, but that makes no sense at all …

a constant of integration should just be added … at the end.

 

[bobmerhebi]

so it should look like this: y2 = -1/6cos23x+ c? I am not getting it! It seems to be the same but a famile of solutions which i think are still invalid. Am I missing something?

 

[tiny-tim]

so it should look like this: y² = -1/6cos²3x+ c?

Yes!

(except personally I'd write it y² = C - 1/6cos²3x)

 

[gabbagabbahey]

What is wrong with taking the square root of a negative number? You just get an imaginary value of y, no big deal...

If you only want real-valued solutions, then you have to restrict your Domain.

 

[bobmerhebi]

in this case the constant should be greater or equal than 1/6cos23x. so here there should be a condition right?

thx 4 the help

 

[gabbagabbahey]

Your constant is a constant...how can it always be larger than $\frac{1}{6cos^2(3x)}$? ??

Instead, if you want only the real-valued solutions, you have to restrict your x-values (your Domain): only values of x for which $\frac{-1}{6cos^2(3x)}+C \geq 0$ give real solutions for y.

Your constant $C$ is determined by the initial conditions of the system. For example; if you are told that y(0)=1/sqrt(6), then C will be 1/3.

 

[tiny-tim]

in this case the constant should be greater or equal than 1/6cos23x …

No, the constant should be ≥ 0.

(and as gabbagabbahey says, x is limited)

 

[bobmerhebi]

aha ok then the independent variable should be over a restricted domain to get real valued solutions otherwise the solution is in the complex system. thank you all 4 ur help