Ordinary Differential Equations!Urgent Help Neede

**noob** · June 13th 2008, 05:42 AM

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

**PaulRS** · June 13th 2008, 05:51 AM

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

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

**noob** · June 13th 2008, 05:55 AM

Originally Posted by PaulRS

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

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

**topsquark** · June 13th 2008, 06:00 AM

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

**kalagota** · June 13th 2008, 06:09 AM

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

if my memory is still good, this might help you..

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

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

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

and start using the method here.

**noob** · June 13th 2008, 07:23 AM

Originally Posted by kalagota

if my memory is still good, this might help you..

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

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

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

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??

**noob** · June 13th 2008, 07:29 AM

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

answer for first questions

**Isomorphism** · June 13th 2008, 07:42 AM

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: $(2xy)' = 2x'y + 2xy' = 2y+2xy'$

So lets differentiate:

$y' + 2xy - x = 0$

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

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

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

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

**noob** · June 13th 2008, 08:52 AM

Originally Posted by Isomorphism

Product rule: $(2xy)' = 2x'y + 2xy' = 2y+2xy'$

So lets differentiate:

$y' + 2xy - x = 0$

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

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

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

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

thanks mates..its really help

**noob** · June 13th 2008, 08:54 AM

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

answer for first questions

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

**Isomorphism** · June 13th 2008, 08:55 AM

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?

**noob** · June 13th 2008, 09:12 AM

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

answer for first questions

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

**Isomorphism** · June 13th 2008, 10:20 AM

Its correct

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

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

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

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

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

**noob** · June 13th 2008, 10:31 AM

Originally Posted by Isomorphism

Its correct

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

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

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

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

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

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

**Isomorphism** · June 13th 2008, 10:34 AM

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..

Thread: Ordinary Differential Equations!Urgent Help Neede

LinkBack

Thread Tools

Display

Ordinary Differential Equations!Urgent Help Neede

Similar Math Help Forum Discussions

Ordinary differential equations

Ordinary Differential Equations

Ordinary differential equations

Ordinary differential equations....

Ordinary Differential Equations

Search Tags