Linear homogeneous differential equations of 2nd order Step-By-Step
For example, you have entered (calculator here): 11y(x)−2dxdy(x)+3dx2d2y(x)=0
Detail solution
Divide both sides of the equation by the multiplier of the derivative of y'': 3 We get the equation: 311y(x)−32dxdy(x)+dx2d2y(x)=0 This differential equation has the form:
y'' + p*y' + q*y = 0,
where p=−32 q=311 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)=0 In this case, the characteristic equation will be: k2−32k+311=0 Detailed solution of the equation - this is a simple quadratic equation The roots of this equation: k1=31−342i k2=31+342i As there are two roots of the characteristic equation, solving the correspondent differential equation looks as follows: y(x)=ek1xC1+ek2xC2 The final answer: y(x)=C1ex(31−342i)+C2ex(31+342i)