Mister Exam

# Differential equation y′′−6y′+9y=0

y() =
y'() =
y''() =
y'''() =
y''''() =

from to

### The solution

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                          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

### Step

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 linear homogeneous second-order differential equation with constant coefficients.
The equation has an easy solution.

### Step

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 $$k_{1} = 3$$
$$y{\left(x \right)} = C_{2} x e^{3 x} + C_{1} e^{3 x}$$
                    3*x
y(x) = (C1 + C2*x)*e   
$$y{\left(x \right)} = \left(C_{2} x + C_{1}\right) e^{3 x}$$
$$y\left(x\right)=x\,e^{3\,x}\,\left(\left.{{d}\over{d\,x}}\,y\left(x \right)\right|_{x=0}\right)-3\,y\left(0\right)\,x\,e^{3\,x}+y\left(0 \right)\,e^{3\,x}$$
y = x*E^(3*x)*('at('diff(y,x,1),x = 0))-3*y(0)*x*E^(3*x)+y(0)*E^(3*x)