# Thread: Ordinary Differential Equations!Urgent Help Neede

1. ## Ordinary Differential Equations!Urgent Help Neede

Hye all, as a newbie in this forum, i would be really sorry if got any mistake or indirectly breaching the terms and condition in this forum.

I'm just really2 need a help from anyone of you, please helping me to attain the solution of both questions. The question is on Ordinary Differential Equations topic. This is my important assignment question as it will bring 15% of my career marks (continuous assessment) before final exam. Please really help as that 15% is too important to me..thanks in advanced mates

2. 1. $\displaystyle y'=3+5y$ thus $\displaystyle \frac{y'}{3+5y}=1$

Now integrate: $\displaystyle \int_0^{x}\frac{y'}{3+5y}dt=x$ (let $\displaystyle u=y(t)$ on the LHS)

3. Originally Posted by PaulRS
1. $\displaystyle y'=3+5y$ thus $\displaystyle \frac{y'}{3+5y}=1$

Now integrate: $\displaystyle \int_0^{x}\frac{y'}{3+5y}dt=x$ (let $\displaystyle u=y(t)$ on the LHS)

thanks..i need to try by my self first and if u dont mind to check my answer after all is it my answer is correct or not

4. Originally Posted by noob
thanks..i need to try by my self first and if u dont mind to check my answer after all is it my answer is correct or not
Okay, post your solutions and we'll look at them. However you can check your own work by simply plugging the solution back through the original equation.

-Dan

5. Originally Posted by noob
Hye all, as a newbie in this forum, i would be really sorry if got any mistake or indirectly breaching the terms and condition in this forum.

I'm just really2 need a help from anyone of you, please helping me to attain the solution of both questions. The question is on Ordinary Differential Equations topic. This is my important assignment question as it will bring 15% of my career marks (continuous assessment) before final exam. Please really help as that 15% is too important to me..thanks in advanced mates

so for the second one, i remembered we use variation of parameter when we were solving higher orders.

so first, differentiate (with respect to $\displaystyle x$): $\displaystyle y' + 2xy = x$

we obtain: $\displaystyle y'' + 2xy' + 2y = 1$

and start using the method here.

6. Originally Posted by kalagota

so for the second one, i remembered we use variation of parameter when we were solving higher orders.

so first, differentiate (with respect to $\displaystyle x$): $\displaystyle y' + 2xy = x$

we obtain: $\displaystyle y'' + 2xy' + 2y = 1$

and start using the method here.

how can we get y'' + 2xy' + 2y = 1????? is it we get y''+ 2y = 1...where did 2xy' came from??

7. Originally Posted by topsquark
Okay, post your solutions and we'll look at them. However you can check your own work by simply plugging the solution back through the original equation.

-Dan
hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

8. Originally Posted by noob
how can we get y'' + 2xy' + 2y = 1????? is it we get y''+ 2y = 1...where did 2xy' came from??
Product rule: $\displaystyle (2xy)' = 2x'y + 2xy' = 2y+2xy'$

So lets differentiate:

$\displaystyle y' + 2xy - x = 0$

$\displaystyle (y' + 2xy - x)' = 0$

$\displaystyle (y')' +(2xy)' - (x)' = 0$

$\displaystyle y'' + 2y + 2xy' - 1 = 0$

$\displaystyle y'' + 2xy' +2y = 1$

9. Originally Posted by Isomorphism
Product rule: $\displaystyle (2xy)' = 2x'y + 2xy' = 2y+2xy'$

So lets differentiate:

$\displaystyle y' + 2xy - x = 0$

$\displaystyle (y' + 2xy - x)' = 0$

$\displaystyle (y')' +(2xy)' - (x)' = 0$

$\displaystyle y'' + 2y + 2xy' - 1 = 0$

$\displaystyle y'' + 2xy' +2y = 1$
thanks mates..its really help

10. Originally Posted by noob
hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

how bout my first solutions..please comment and add up some ideas if it wrong thanks very2 much

11. Originally Posted by noob
how bout my first solutions..please comment and add up some ideas if it wrong thanks very2 much
I cant read it,why dont you type it out in a post?

12. Originally Posted by noob
hye topsquark, hope u really can help (or the others) this is my answer n im sure i tottally wrong plz check mates..im in my first year of studies and got a bit problem with my maths

Originally Posted by Isomorphism
I cant read it,why dont you type it out in a post?
this is my answer and surely it's wrong because im also did not really sure what am i doing, please refer to my first post regarding the question by the way thanks for your willingness

p/s: really like your siggy..so meaningful

13. Its correct

I cant read the last two lines but it must be:

$\displaystyle 5y = e^{5x+5C} - 3$

$\displaystyle y = \frac15 e^{5x+5C} - \frac35$

However I would add a point that $\displaystyle e^{5C}$ is a constant, so call it A.Thus:

$\displaystyle y = \frac{A}5 e^{5x} - \frac35$

14. Originally Posted by Isomorphism
Its correct

I cant read the last two lines but it must be:

$\displaystyle 5y = e^{5x+5C} - 3$

$\displaystyle y = \frac15 e^{5x+5C} - \frac35$

However I would add a point that $\displaystyle e^{5C}$ is a constant, so call it A.Thus:

$\displaystyle y = \frac{A}5 e^{5x} - \frac35$

thank you very much, from the question needed, is it i can left the answer just like this??one more things that quite confusing me is when did we need to consider the integral factor?? some question in book using an integral factor method

15. Originally Posted by noob
thank you very much, from the question needed, is it i can left the answer just like this??
Yes you can.

one more things that quite confusing me is when did we need to consider the integral factor?? some question in book using an integral factor method
You mean "integrating factor"? There is no need for it here, since the differential equation is in separable form..

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