1. ## differentiation

Solve the differential equation

ydy/dx =x+xy^2 given that y = 0 when x = 0.

i can get down to here but then i get stuck

dy/dx= x(1+y^2)/y

∫▒y/(1+y^2 ) dy=∫▒ x dx i dont know how to integrate this?

2. ## Re: differentiation

The differential equation:
$$\frac{dy}{dx}=x \frac{1+y^2}{y}$$
is of variable seperable type reducing to:
$$\int x\;dx=\int\frac{y}{1+y^2}\;dy$$
which integrates to:

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

3. ## Re: differentiation

How do you integrate the dy

4. ## Re: differentiation

Originally Posted by markosheehan
How do you integrate the dy
substitute

$u=y^2$

$du=2y~dy$

you get

$\dfrac 1 2 \int \dfrac {du}{1+u}$

5. ## Re: differentiation

Originally Posted by romsek
substitute

$u=y^2$

$du=2y~dy$

you get

$\dfrac 1 2 \int \dfrac {du}{1+u}$
A slightly easier substitution would be

$u=y^2+1$

$du=2y~dy$

you get

$\dfrac 1 2 \int \dfrac {du}{u}$

Once you have done the integration for $u$ then switch back to $y$

6. ## Re: differentiation

You recognise it as of the form
$$\frac{1}{2} \int \frac{f'(y)}{f(y)}dy=\frac{1}{2}\int \left( \frac{d}{dy}\ln(f(y)) \right) \;dy=\frac{1}{2} \ln(f(y))$$

7. ## Re: differentiation

Originally Posted by markosheehan
Solve the differential equation

ydy/dx =x+xy^2 given that y = 0 when x = 0.

i can get down to here but then i get stuck

dy/dx= x(1+y^2)/y

∫▒y/(1+y^2 ) dy=∫▒ x dx i dont know how to integrate this?
It can easily be solved by using integration by substitution
$\int {\frac{y}{1+y^{2}}dy}= \int{x dx}$
Substitute $1+y^{2} = u$
$\implies{ydy = \frac{du}{2}}$
$\int{\frac{du}{2u}}= \int{x dx}$
$\frac{1}{2}lnu=\frac{x^{2}}{2}+C$
Put the value of u
$\frac{1}{2}ln(1+y^{2})=\frac{x^{2}}{2}+C$
For more practice of integartion by substitution http://www.actucation.com/calculus-2...y-substitution