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Linear homogeneous differential equations of 2nd order Step-By-Step

For example, you have entered (calculator here):
11y(x)−2ddxy(x)+3d2dx2y(x)=011 y{\left(x \right)} - 2 \frac{d}{d x} y{\left(x \right)} + 3 \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 0

Detail solution

Divide both sides of the equation by the multiplier of the derivative of y'':
33
We get the equation:
11y(x)3−2ddxy(x)3+d2dx2y(x)=0\frac{11 y{\left(x \right)}}{3} - \frac{2 \frac{d}{d x} y{\left(x \right)}}{3} + \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=−23p = - \frac{2}{3}
q=113q = \frac{11}{3}
It is called linear homogeneous
second-order differential equation with constant coefficients.
The equation has an easy solution
We solve the corresponding homogeneous linear equation
y'' + p*y' + q*y = 0

First of all we should find the roots of the characteristic equation
q+(k2+kp)=0q + \left(k^{2} + k p\right) = 0
In this case, the characteristic equation will be:
k2−2k3+113=0k^{2} - \frac{2 k}{3} + \frac{11}{3} = 0
Detailed solution of the equation
- this is a simple quadratic equation
The roots of this equation:
k1=13−42i3k_{1} = \frac{1}{3} - \frac{4 \sqrt{2} i}{3}
k2=13+42i3k_{2} = \frac{1}{3} + \frac{4 \sqrt{2} i}{3}
As there are two roots of the characteristic equation,
solving the correspondent differential equation looks as follows:
y(x)=ek1xC1+ek2xC2y{\left(x \right)} = e^{k_{1} x} C_{1} + e^{k_{2} x} C_{2}
The final answer:
y(x)=C1ex(13−42i3)+C2ex(13+42i3)y{\left(x \right)} = C_{1} e^{x \left(\frac{1}{3} - \frac{4 \sqrt{2} i}{3}\right)} + C_{2} e^{x \left(\frac{1}{3} + \frac{4 \sqrt{2} i}{3}\right)}
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