# Thread: [SOLVED] Particular solution of a diffy eq

1. ## [SOLVED] Particular solution of a diffy eq

Hi

There's this differential equation :
$\cosh(x) y'-\sinh(x) y=\sinh(x)$
Where $y'=\frac{dy}{dx}$

The general solution is $y=k \cosh(x)+y_0$
where $y_0$ is a defined function, solution of the equation.
But I can't find it

2. Originally Posted by Moo
Hi

There's this differential equation :
$\cosh(x) y'-\sinh(x) y=\sinh(x)$
Where $y'=\frac{dy}{dx}$

The general solution is $y=k \cosh(x)+y_0$
where $y_0$ is a defined function, solution of the equation.
But I can't find it
Divide everything through by $\cosh{x}$ and then it is first order linear, which you can solve using the integrating factor method.

Note: $\int{\tanh{x}\,dx} = \log{\cosh{x}}$.

3. Originally Posted by Prove It
Divide everything through by $\cosh{x}$ and then it is first order linear, which you can solve using the integrating factor method.

Note: $\int{\tanh{x}\,dx} = \log{\cosh{x}}$.
$y'=\tanh(x) y+\sinh(x)$

But I don't see how to solve it.

Actually, this is how I found the general solution :
we know that if we substitute by $z=y-y_0$, where $y_0$ is a particular solution, we get :
$\cosh(x) z'-\sinh(x) z=0$
This clearly gives $z=k \cosh(x)$ (using this integrating factor)

So the general solution is $y=k \cosh(x)+y_0$.

But the problem is to find $y_0$ ><

4. Originally Posted by Moo
$y'=\tanh(x) y+\sinh(x)$

But I don't see how to solve it.

Actually, this is how I found the general solution :
we know that if we substitute by $z=y-y_0$, where $y_0$ is a particular solution, we get :
$\cosh(x) z'-\sinh(x) z=0$
This clearly gives $z=k \cosh(x)$ (using this integrating factor)

So the general solution is $y=k \cosh(x)+y_0$.

But the problem is to find $y_0$ ><
$\frac{dy}{dx} -\tanh{(x)}y = \tanh{x}$.

The integrating factor is $e^{\int{-\tanh{x}\,dx}} = \frac{1}{\cosh{x}}$ (Check).

So using the integrating factor method

$\frac{d}{dx}\left(\frac{y}{\cosh{x}}\right) = \frac{\sinh{x}}{\cosh^2{x}}$.

Can you go from here?

5. Originally Posted by Prove It
$\frac{dy}{dx} -\tanh{(x)}y = \tanh{x}$.

The integrating factor is $e^{\int{-\tanh{x}\,dx}} = \frac{1}{\cosh{x}}$ (Check).

So using the integrating factor method

$\frac{d}{dx}\left(\frac{y}{\cosh{x}}\right) = \frac{\sinh{x}}{\cosh^2{x}}$.

Can you go from here?
if we solve for y in the latter expression, we should get y=-1 ? oO

What a fool... Thanks ^^

but though, can you explain how to get the latter expression ?
do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)

6. Originally Posted by Moo
if we solve for y in the latter expression, we should get y=-1 ? oO

What a fool... Thanks ^^

but though, can you explain how to get the latter expression ?
do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)
Yes I get y = -1 + C.

Integrating factor - Wikipedia, the free encyclopedia

Hang on...

7. Originally Posted by Prove It
Yes I get y = -1 + C.

Integrating factor - Wikipedia, the free encyclopedia

Hang on...
Try here

8. Originally Posted by Prove It
Thanks

9. Originally Posted by Moo
Hi

There's this differential equation :
$\cosh(x) y'-\sinh(x) y=\sinh(x)$

Let $y=f(x)$ be a solution.
Then it means, $\cosh (x) f ' (x) - \sinh (x) f( x) = \sinh (x)$ for all $x\in \mathbb{R}$.
Thus, $\cosh (-x) f ' (-x) - \sinh (-x) f( -x) = \sinh (-x)$ for all $x\in \mathbb{R}$.
Thus, $\cosh (x) f ' (-x) + \sinh (x) f( - x) = - \sinh (x)$ for all $x\in \mathbb{R}$.

What have we established, Moo?
We have shown that if $f(x)$ solves $\cosh (x) y' - \sinh (x) y = \sinh (x)$
Then $f(-x)$ solves $\cosh (x) y' + \sinh (x) y = -\sinh (x)$.

Now, $\cosh (x) y' + \sinh (x) y = - \sinh (x) \implies [ \sinh (x) y] ' = - \sinh (x)$ [1]

Can you take it from there?
(When you solve [1] it would mean $y(-x)$ solves the original problem).

10. Originally Posted by Moo
do you know a website that explains the methods of solving d.e. ? (apart from wikipedia)
Ummmm...my tutorial!!!!