# Thread: General solution of a Second order differential equations

1. ## General solution of a Second order differential equations

Hi,

Heres the question:

Find the general solution of the following ordinary differential equation

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

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

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

so $\displaystyle 5kx +4k = x$

How do i solve the above

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

4. 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...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. EDIT: I'm an idiot

6. 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 $\displaystyle x^n$ on the right side, you will need to try a polynomial of the form $\displaystyle Ax^n+ Bx^{n-1}+ \cdot\cdot\cdot+ Yx+ Z$- that is all powers up to n.