2nd order nonhomo DE

**zangestu888** · June 15th 2009, 08:29 PM

Am not sure at all how to do this, all i know is the intial guess the DE is

y''+ 5y'=30x^2

my intial guess is ax^2 +bx + c

the deritives are y'=2ax + b and y''=2a

then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
10ax + 2a + 5b=30x^2
if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!

**Sampras** · June 15th 2009, 08:58 PM

Originally Posted by zangestu888

Am not sure at all how to do this, all i know is the intial guess the DE is

y''+ 5y'=30x^2

my intial guess is ax^2 +bx + c

the deritives are y'=2ax + b and y''=2a

then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
10ax + 2a + 5b=30x^2
if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!

Let $\mathcal{L} \{y(x) \} = Y(s)$ . Take the Laplace transform of both sides. So $\mathcal{L} \{y''+5y' \} = \mathcal{L} \{30x^2 \}$ . Solve for $Y(s)$ . Then take inverse Laplace transforms to get $y(x)$ .

**mr fantastic** · June 15th 2009, 11:02 PM

Originally Posted by zangestu888

Am not sure at all how to do this, all i know is the intial guess the DE is

y''+ 5y'=30x^2

my intial guess is ax^2 +bx + c

the deritives are y'=2ax + b and y''=2a

then i get (2a) + 5(2ax+b)=30x^2 or 2a + 10ax + 5b=30x^2
10ax + 2a + 5b=30x^2
if i equate them dont i get a=b=0? or am i missing something any help appreciated thanks in advance!

My post in this thread: http://www.mathhelpforum.com/math-he...-integral.html

suggests the modification you need to make to your particular solution.

Originally Posted by Sampras

Let $\mathcal{L} \{y(x) \} = Y(s)$ . Take the Laplace transform of both sides. So $\mathcal{L} \{y''+5y' \} = \mathcal{L} \{30x^2 \}$ . Solve for $Y(s)$ . Then take inverse Laplace transforms to get $y(x)$ .

From the work posted by the OP I think it's clear that this technique is probably beyond what s/he has learned so far.

**zangestu888** · June 16th 2009, 07:54 AM

thanks for ur help but we havent startd this laplace transformation, were doing undertermined coeefcients now, any other way? are my coeefficients correct.

**HallsofIvy** · June 16th 2009, 09:06 AM

The characteristic equation for this $r^2+ 5r= 0$ which has roots 0 and -5 so the homogenous equation has solutions $C_1e^{-5x}$ and $C_2e^{-x}= C_2$ , a constant. Since $30x^2= 30x^2e^{0x}$ is of that "type", you need to multiply by "x". Try $y= x(ax^2 +bx + c)= ax^3+ bx^2+ cx$

**mr fantastic** · June 16th 2009, 07:41 PM

Originally Posted by zangestu888

thanks for ur help but we havent startd this laplace transformation, were doing undertermined coeefcients now, any other way? are my coeefficients correct.

Did you read my post in the thread whose link I gave you?

**zangestu888** · June 16th 2009, 07:46 PM

o noo thanks! il look at it now!! thanks!

**zangestu888** · June 16th 2009, 07:50 PM

okay ithink i got it am gonna redo it since $e^0$ is actually in the homogenous side of the DE i must do a multiple of x k thanks am gonna work on it now, a question i can use variation of paramters?

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