# Thread: DE IVP seperable variable

1. ## DE IVP seperable variable

Hi again
subject:first order linear DE
Using variable separable method,
I have no idea on how to start on these 3 problems, please guide me.
a)dy/dx=(2-y)(2+y),y(0)=-1
b)dy/dx=y(y-1),y(0)=2
c)dy/dx=(y-2)(y-4),y(0)=3

are they solve using the same method? so even if y move to other place,it does not change the method of solving?Thanks in advance

2. oops it first order DE not linear DE

3. a) $\displaystyle \displaystyle \frac{1}{(2-y)(2+y)}\,\frac{dy}{dx} = 1$. Integrate both sides with respect to $\displaystyle \displaystyle x$.

b) $\displaystyle \displaystyle \frac{1}{y(y-1)}\,\frac{dy}{dx} = 1$. Integrate both sides with respect to $\displaystyle \displaystyle x$.

c) $\displaystyle \displaystyle \frac{1}{(y-2)(y - 4)}\,\frac{dy}{dx} = 1$. Integrate both sides with respect to $\displaystyle \displaystyle x$.

4. i know it to do that but how to integrate the LHS, lets take a for example, it becomes ln|(2-y)(2+y)| or ln(4+y^2)?
so all 3 uses the same method to solve?if so could you show me 1 example, then i will use it as base for the other 2, thanks in advance

5. E.g. in the first one you should get to

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

You need to use Partial Fractions to evaluate the LHS.

6. for case a) i manage to get A= 1/4 and B= 1/4 then i stuck again,how to continue from there?

7. Originally Posted by salohcinseah
for case a) i manage to get A= 1/4 and B= 1/4 then i stuck again,how to continue from there?
I guess you mean you have $\displaystyle \frac{1}{(2- y)(2+ y)}= \frac{\frac{1}{4}}{2- y}+ \frac{\frac{1}{4}}{2+ y}$
Now, you need to integrate
$\displaystyle \frac{1}{4}\int \frac{dy}{2- y}+ \frac{1}{4}\int\frac{dy}{2+ y}$
If you cannot see how to do that immediately, use substitutions- in the first let u= 2- y and in the second, u= 2+ y.

May I ask what level course this is? Integration by partial fractions and substitution usually come fairly early in a Calculus course- certainly earlier than solving differential equations!

8. it basic calculus for university it for engineering degree the thing is ,it sometimes very confusing,these questions are from DE so when i intergrate it it become 1/4 ln|2-y|+1/4 ln |2+y| = x + c then how i can do so ithat i can make c the subject to find it value?

9. Substitute the values given in your initial condition...

10. y(0)=-1 but there 2 ln and 2 y and 2 1/4 in the LHS that y i have no idea on how to my it into y= something something +c could you help me by solving it step by step so i can use it as a base to solve the other 2?

11. You know that when $\displaystyle \displaystyle x = 0, y = -1$. So wherever you see an $\displaystyle \displaystyle x$, substitute $\displaystyle \displaystyle 0$, and wherever you see a $\displaystyle \displaystyle y$, substitute $\displaystyle \displaystyle -1$.

12. arh stupid me, i was thinking of bring all y to RHS so if i put y as -1, all i need to do is bring x to LHS thus i get c, am i right?

13. Yes. You can probably also use logarithm laws to simplify...

14. thanks to all whom have helped me