⌨

You have entered
[src]

2 d d - 6*--(y(x)) + 9*y(x) + ---(y(x)) = 0 dx 2 dx

$$9 y{\left(x \right)} - 6 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0$$

9*y - 6*y' + y'' = 0

Detail solution

Given the equation:

$$9 y{\left(x \right)} - 6 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0$$

This differential equation has the form:

$y'' + p*y' + q*y = 0$,

where

$$p = -6$$

$$q = 9$$

It is called

The equation has an easy solution.

First of all, we should solve the correspondent linear homogeneous equation

$$y'' + p*y' + q*y = 0$$

First of all we should find the roots of the characteristic equation:

$$k^{2} + k p + q = 0$$

In this case, the characteristic equation will be:

$$k^{2} - 6 k + 9 = 0$$

Detailed solution of the equation

- this is a simple quadratic equation.

The root of this equation:

$$k_{1} = 3$$

As there is one root of the characteristic equation,

and it is not complex, then

solving the correspondent differential equation looks as follows:

$$y{\left(x \right)} = C_{1} e^{k_{1} x} + C_{2} x e^{k_{1} x}$$

Substitute $$k_{1} = 3$$

The final answer:

$$y{\left(x \right)} = C_{2} x e^{3 x} + C_{1} e^{3 x}$$

The answer
[src]

3*x y(x) = (C1 + C2*x)*e

$$y{\left(x \right)} = \left(C_{2} x + C_{1}\right) e^{3 x}$$

The answer (#2)
[src]

$$y\left(x\right)=x\,e^{3\,x}\,\left(\left.{{d}\over{d\,x}}\,y\left(x
\right)\right|_{x=0}\right)-3\,y\left(0\right)\,x\,e^{3\,x}+y\left(0
\right)\,e^{3\,x}$$

y = x*E^(3*x)*('at('diff(y,x,1),x = 0))-3*y(0)*x*E^(3*x)+y(0)*E^(3*x)

The classification

2nd power series ordinary

factorable

nth linear constant coeff homogeneous