Separable Differential Equations Problems

**Paymemoney** · August 28th 2010, 02:10 AM

Hi
The following questions i need help on:
Solve the Differential Equations
1) $\frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0$

$\frac{dy}{dx} = \frac{-3x^2}{(1+x^2)}*\frac{x}{y}$

$\frac{dy}{dx} = \frac{-3x^2}{x(1+x^2)}$

$\frac{dy}{dx} = \frac{-3x^2y}{x+x^3)}$

$\frac{dy}{dx} = \frac{x^2(-3y}{x(1+x^2)}$

$\frac{dy}{dx} = \frac{x(-3y)}{(1+x^2)}$

$\frac{1}{-3y} * \frac{dy}{dx} = \frac{x}{(1+x^2)}$

$u=1+x^2$
$du=2xdx$

$\frac{-1}{3}ln|y| = \int \frac{1}{u} * \frac{du}{2}$

$\frac{-1}{3}ln|y| = \frac{1}{2} \int \frac{1}{u}$

$\frac{-1}{3}ln|y| = \frac{1}{2}ln|x^2+1|$

$-ln|y| = ln|(x^2+1)^\frac{3}{2}|$

$y = -e^{ln|(x^2+1)^\frac{3}{2}|+C}$

$A=-e^C$

$y = Ae^{ln|(x^2+1)^\frac{3}{2}|}$

2) $\frac{dy}{dx} = y^2cos2x$

$\int \frac{1}{y^2} dy = \int cos2xdx$

u=2x du=2 dx

$\frac{-1}{y} = \frac{1}{2} \int cos(u) du$

$\frac{-1}{y} = \frac{1}{2}sin(u) + C$

$\frac{-2}{y} = sin(u) + C$

$y = \frac{-2}{sin2x + C}$

3) $\frac{dy}{dx} =\frac{y^2-1}{x}$

$\int \frac{1}{y^2-1} dy = \int \frac{1}{x} dx$

what should i do next?

i tried $coth^{-1}$ but that didn't work.

P.S

**mathaddict** · August 28th 2010, 02:18 AM

Originally Posted by Paymemoney

Hi
The following questions i need help on:
Solve the Differential Equations
1) $\frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0$

$\frac{dy}{dx} = \frac{-3x^2}{(1+x^2)}*\frac{x}{y}$

$\frac{dy}{dx} = \frac{-3x^2}{x(1+x^2)}$

$\frac{dy}{dx} = \frac{-3x^2y}{x+x^3)}$

$\frac{dy}{dx} = \frac{x^2(-3y}{x(1+x^2)}$

$\frac{dy}{dx} = \frac{x(-3y)}{(1+x^2)}$

$\frac{1}{-3y} * \frac{dy}{dx} = \frac{x}{(1+x^2)}$

$u=1+x^2$
$du=2xdx$

$\frac{-1}{3}ln|y| = \int \frac{1}{u} * \frac{du}{2}$

$\frac{-1}{3}ln|y| = \frac{1}{2} \int \frac{1}{u}$

$\frac{-1}{3}ln|y| = \frac{1}{2}ln|x^2+1|$

$-ln|y| = ln|(x^2+1)^\frac{3}{2}|$

$y = -e^{ln|(x^2+1)^\frac{3}{2}|+C}$

$A=-e^C$

$y = Ae^{ln|(x^2+1)^\frac{3}{2}|}$

2) $\frac{dy}{dx} = y^2cos2x$

$\int \frac{1}{y^2} dy = \int cos2xdx$

u=2x du=2 dx

$\frac{-1}{y} = \frac{1}{2} \int cos(u) du$

$\frac{-1}{y} = \frac{1}{2}sin(u) + C$

$\frac{-2}{y} = sin(u) + C$

$y = \frac{-2}{sin2x + C}$

3) $\frac{dy}{dx} =\frac{y^2-1}{x}$

$\int \frac{1}{y^2-1} dy = \int \frac{1}{x} dx$

what should i do next?

i tried $coth^{-1}$ but that didn't work.

P.S

For(3), let $y=\sin \theta$

Are you having problems with the rest of the questions or you want to verify the answers?

**Paymemoney** · August 28th 2010, 03:48 AM

question 1) and 2) are incorrect according to the book's answers.

The book's answers say the following:

1) $y= \frac{C}{(x+x^2)^{\frac{3}{2}}}$

2) $y=\frac{2}{C-sin(2x)}$

**Paymemoney** · August 29th 2010, 11:48 PM

can anyone help me??

**raj007** · August 30th 2010, 01:33 AM

$\int(\frac{dy}{y^2 -1}) = \int(\frac{dx}{x})$

$\int(\frac{\frac{y + 1}{2} - \frac{(y - 1)}{2}}{(y + 1)(y - 1)})dy = log|x| + log|c|$

$\frac{log(y - 1) - log(y + 1)}{2} = log|cx|\\$

$log|\frac{y - 1}{y + 1}| = log|x^2.k|$

so

$\frac{k(y-1)}{(y + 1)} = x^2$

**Ackbeet** · August 30th 2010, 02:46 AM

Your algebra for (1) is incorrect. As far as I can tell, your mistake was in going from step 1 to step 2. Try this:

$\displaystyle{\frac{x}{y} * \frac{dy}{dx} + \frac{3x^2}{(1+x^2)}=0}$

$\displaystyle{\frac{x}{y} * \frac{dy}{dx} =- \frac{3x^2}{(1+x^2)}}$

$\displaystyle{\frac{dy}{y}=-\frac{3x}{(1+x^2)}\,dx}.$

Integrate, etc.

As for the second problem, your answer and the book's answer are equivalent. You've just re-defined your constant of integration once or twice. Your constant is the negative of the book's constant. That is an unimportant difference.

As for #3, raj007's solution is a bit bizarre, in my opinion. I would go with a straight-forward trig substitution, or look the integral up in a table if you're allowed to do that.

**Ackbeet** · August 30th 2010, 05:46 AM

Concerning Problem #3: My idea of trig substitution is not, I think, useful. Try partial fractions after factoring the denominator on the LHS.

**Paymemoney** · August 30th 2010, 03:32 PM

$y=sin\theta dy=cos\theta d\theta$

$\int \frac{1}{sin^2(\theta)-1} * cos\theta d\theta = ln|x|+C$

$\int \frac{cos\theta}{sin^2(\theta)-1} = ln|x|+C$

$\int \frac{1}{cos^2(\theta)} = ln|x|+C$

$\int sec^2(\theta)=ln|x|+C$

What should i do next?

**Defunkt** · August 30th 2010, 04:20 PM

The cosine in your numerator dissappeared. A minus in the denominator has also gone astray.

And a shorter way will be to factor $y^2 -1$ into $(y+1)(y-1)$ and then do partial fractions.

**Paymemoney** · August 30th 2010, 08:27 PM

$\int \frac{1}{(y-1)(y+1)} dy = \int \frac{1}{x} dx$

$\frac{-1}{2}\int\frac{1}{y-1} + \frac{1}{2} \int \frac{1}{y+1}dy = ln|x|+C$

$\frac{-1}{2}ln|y-1 + \frac{1}{2} ln|y+1| = ln|x|+C$

$\frac{-1}{2}(ln|y-1|-ln|y+1|) = ln|x|+C$

$\frac{-1}{2}ln|\frac{y-1}{y+1}| = ln|x|+C$

what should be done next??

**mathaddict** · August 30th 2010, 08:48 PM

Originally Posted by Paymemoney

$\int \frac{1}{(y-1)(y+1)} dy = \int \frac{1}{x} dx$

$\frac{-1}{2}\int\frac{1}{y-1} + \frac{1}{2} \int \frac{1}{y+1}dy = ln|x|+C$

$\frac{-1}{2}ln|y-1 + \frac{1}{2} ln|y+1| = ln|x|+C$

$\frac{-1}{2}(ln|y-1|-ln|y+1|) = ln|x|+C$

$\frac{-1}{2}ln|\frac{y-1}{y+1}| = ln|x|+C$

what should be done next??

From step 3 to step 4, how did the + becomes - ?

Next, put the equation in exponents form.

**Paymemoney** · August 31st 2010, 01:41 AM

oops my mistake it's meant to be:
$<br /> \frac{1}{2}(ln|y+1|-ln|y-1|) = ln|x|+C$

$ln|\frac{y+1}{y-1}|=2ln|x|+2C$

$\frac{y+1}{y-1}=e^{ln|x^2|}*e^{2C}$

$y+1=x^2*e^{2C}$
$y+1=(x^2*e^{2C})(y-1)$
$y+1=x^2*e^{2C}y-x^2*e^{2C}$
$1+x^2*e^{2C}=x^2*e^{2C}y-y$
$1+x^2*e^{2C}=y(x^2*e^{2C}-1)$

$y=\frac{1+x^2*e^{2C}}{-1+x^2*e^{2C}}$

**Paymemoney** · August 31st 2010, 02:03 AM

Originally Posted by Paymemoney

$y=\frac{1+x^2*e^{2C}}{-1+x^2*e^{2C}}$

However answers are $\frac{1+e^{2C}x^2}{1-e^{2C}x^2}$

**Ackbeet** · August 31st 2010, 02:06 AM

That looks good, except that in going from step 3 to 4, you lost your y-1 factor. And then in going from step 4 to step 5, you regained it! I wouldn't recommend dropping terms willy-nilly, unless you have a good reason for doing it (approximation, for example).

Recommendation: rename your constant of integration so it looks a bit cleaner. That is, define, say, K = exp(2C), and use K in both places there. This is not a fundamentally important aspect of the problem, however. It is merely cosmetic.

**Ackbeet** · August 31st 2010, 02:11 AM

To fix the sign error, I'd go back and re-visit your partial fraction decomposition. I think you might have those signs flipped.

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