# Thread: Separable Differential Equations Problems

1. ## Separable Differential Equations Problems

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

2. 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?

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

4. can anyone help me??

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

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

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

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

so

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

6. 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.

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

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

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

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

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

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

What should i do next?

9. 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.

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

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

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

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

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

what should be done next??

11. 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.

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

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

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

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

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

13. Originally Posted by Paymemoney
$y=\frac{1+x^2*e^{2C}}{-1+x^2*e^{2C}}$
However answers are $\frac{1+e^{2C}x^2}{1-e^{2C}x^2}$

14. 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.

15. 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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