General solution of a Second order differential equations

**Beard** · March 3rd 2010, 10:27 AM

Hi,

Heres the question:

Find the general solution of the following ordinary differential equation

$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = x$

So far I have found out the complementary function but I can't seem to work out the partial function.

$y_{pi} = kx, \frac{dy_{pi}}{dx} = k , \frac{d^2y_{pi}}{dx^2} = 0$

so $5kx +4k = x$

How do i solve the above

**chisigma** · March 3rd 2010, 11:02 AM

First we find the general integral of the 'incomplete equation' solving the second order equation...

$s^{2} + 4s + 5 = 0$ (1)

... that has the solutions $s= -2 \pm i$ . The general integral of the 'incomplete equation' is then...

$y_{g} (x) = e^{-2x}\cdot (c_{1}\cdot \cos x + c_{2}\cdot \sin x)$ (2)

The next step is to find a particular integral of the 'complete equation'. At this purpose we search a function of the type $y_{p} (x) = a\cdot x + b$ . If we substitute $y_{p} (x)$ in the DE we find $a= \frac{1}{5}$ and $b= - \frac{4}{25}$ . The general integral of the 'complete equation' is then...

$y(x) = y_{g} (x) + y_{p} (x) = e^{-2x}\cdot (c_{1}\cdot \cos x + c_{2}\cdot \sin x) + \frac{1}{5}\cdot x -\frac{4}{25}$ (3)

Kind regards

$\chi$ $\sigma$

**Beard** · March 3rd 2010, 11:28 AM

I got the general equation as the same but where did you manage to get the a= 1/5 and b = -4/25 from?

**chisigma** · March 3rd 2010, 11:40 AM

Originally Posted by Beard

I got the general equation as the same but where did you manage to get the a= 1/5 and b = -4/25 from?

The DE is...

$y^{''} + 4 y^{'} + 5 y = x$ (1)

... and we search a particular solution of the form...

$y = a\cdot x + b \rightarrow y^{'}= a , y^{''}=0$ (2)

Now subsituting (2) in (1) we obtain...

$5ax + 5b + 4a = x$ (3)

... so that is $a = \frac{1}{5}$ , $b = - \frac{4}{25}$ ...

Kind regards

$\chi$ $\sigma$

**Beard** · March 3rd 2010, 12:07 PM

EDIT: I'm an idiot

**HallsofIvy** · March 4th 2010, 05:58 AM

Originally Posted by Beard

EDIT: I'm an idiot

Well, I can't speak to that but it is generally true that when you have $x^n$ on the right side, you will need to try a polynomial of the form $Ax^n+ Bx^{n-1}+ \cdot\cdot\cdot+ Yx+ Z$ - that is all powers up to n.

Thread: General solution of a Second order differential equations

LinkBack

Thread Tools

Display

General solution of a Second order differential equations

Similar Math Help Forum Discussions

[SOLVED] Differential Equations General solution help.

Finding the General Solution of a 2nd Order Linear Ordinary Differential Equation

Differential Equations General Solution

general solution to linear first order differential equations

general solution for differential equations

Search Tags