# Homework Help: Plz help in the result of: sin3xdx+2y(cos3x)^3dy =0

1. Nov 4, 2008

### 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.

2. Nov 4, 2008

### 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?

3. Nov 4, 2008

### HallsofIvy

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

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

4. Nov 4, 2008

### 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.

5. Nov 4, 2008

### bobmerhebi

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

6. Nov 4, 2008

### tiny-tim

Welcome to PF!

Hi bobmerhebi! Welcome to PF!
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"
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?

7. Nov 4, 2008

### bobmerhebi

Re: Welcome to PF!

8. Nov 4, 2008

### tiny-tim

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

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

9. Nov 4, 2008

### bobmerhebi

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

10. Nov 4, 2008

### tiny-tim

Yes!

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

11. Nov 4, 2008

### 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.

12. Nov 4, 2008

### 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

13. Nov 4, 2008

### 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.

Last edited: Nov 4, 2008
14. Nov 4, 2008

### tiny-tim

No, the constant should be ≥ 0.

(and as gabbagabbahey says, x is limited)

15. Nov 4, 2008

### 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