could you please help me to solve thease 4 DEs :
1. i've integrated the last function with maple but the result was very complicated and an answer in my book was simple, but i can't see it.
2.how can i get the other solutions? in the answers there was mentioned something with ln.
3. what i am doing wrong, because answer in the book is something else, like sin and sh functions.
4.what should i do next?

2. Originally Posted by noteiler
could you please help me to solve thease 4 DEs :
1. i've integrated the last function with maple but the result was very complicated and an answer in my book was simple, but i can't see it.
2.how can i get the other solutions? in the answers there was mentioned something with ln.
3. what i am doing wrong, because answer in the book is something else, like sin and sh functions.
4.what should i do next?

$\sqrt{C_1 - \frac{1}{y^2}} = \frac{\sqrt{C_1 y^2 - 1}}{y}$ (assuming y > 0). So the substitution $u = C_1 y^2 - 1$ is clearly suggested.

3. just hit me how to solve 4.
$C1y=y'/(1-y')$

4. For the second...

$\frac{d^2y}{dx^2} = 2y\,\frac{dy}{dx}$

$\int{\frac{d^2y}{dx^2}\,dx} = \int{2y\,\frac{dy}{dx}\,dx}$

$\int{\frac{d^2y}{dx^2}\,dx} = \int{2y\,dy}$

$\frac{dy}{dx} = y^2 + C$.

Now this is a separable DE

$\frac{1}{y^2 + C}\,\frac{dy}{dx} = 1$

$\int{\frac{1}{y^2 + C}\,\frac{dy}{dx}\,dx} = \int{1\,dx}$

$\int{\frac{1}{y^2 + C}\,dy} = \int{1\,dx}$.

The RHS is easy to evaluate. The LHS requires trig substitution.

Let $y = \sqrt{C}\tan{\theta}$ so that $dy = \sqrt{C}\sec^2{\theta}\,d\theta$.

Note that $\theta = \arctan{\frac{y}{\sqrt{C}}}$

LHS becomes

$\int{\frac{1}{y^2 + C}\,dy} = \int{\frac{1}{(\sqrt{C}\tan{\theta})^2 + C}\,\sqrt{C}\sec^2{\theta}\,d\theta}$

$= \int{\frac{\sqrt{C}\sec^2{\theta}}{C\tan^2{\theta} + C}\,d\theta}$

$= \int{\frac{\sqrt{C}\sec^2{\theta}}{C(\tan^2{\theta } + 1)}\,d\theta}$

$= \int{\frac{\sqrt{C}\sec^2{\theta}}{C\sec^2{\theta} }\,d\theta}$

$= \int{\frac{\sqrt{C}}{C}\,d\theta}$

$= \frac{\sqrt{C}}{C}\theta$

$= \frac{\sqrt{C}}{C}\arctan{\frac{y}{\sqrt{C}}}$.

So that means...

$\int{\frac{1}{y^2 + C}\,dy} = \int{1\,dx}$

$\frac{\sqrt{C}}{C}\arctan{\frac{y}{\sqrt{C}}} = x + D$

$\arctan{\frac{y}{\sqrt{C}}} = \sqrt{C}(x + D)$

$\frac{y}{\sqrt{C}} = \tan{[\sqrt{C}(x + D)]}$

$y = \sqrt{C}\tan{[\sqrt{C}(x + D)]}$.

Now you can use your boundary conditions to solve for $C$ and $D$.

5. Originally Posted by noteiler
just hit me how to solve 4.
$C1y=y'/(1-y')$
Note that $\frac{p}{1-p} = -1 + \frac{1}{1-p}$.