 # Using D'Alembert's Procedure to Find a Second Solution to a 2nd Order Differential Equation when one solution is already known/given

### Concept:

Given a 2nd order differential equation of the form:

$$y''+py'+qy=0$$

...and where one solution to the differential equation ,(y1), is known,

...there exists a function (v) where:

$$v' = e^{-\int{}{}(\frac{2y'_1}{y_1}+p)dx}$$

...such that a second solution to the differential equation (y2) is:

$$y_2 = vy_1$$

...and a general solution to the differential equation is represented by:

$$y_g = C_1y_1+ C_2y_2$$

### (Ex 1): Use D'Alembert's method to find a second solution to the following:

$$x^2y''+3xy+y=0$$

where one solution (y1) is known:

$$y_1 = \frac{1}{x}$$

First put the the equation in the correct form by dividing through by x^2:

$$y''+\frac{3}{x}y'+\frac{y}{x^2}=0$$

Now use the above described method to find a function v'

$$v' = e^{-\int{}{}(\frac{2y'_1}{y_1}+p)dx}$$

...where (for our problem):

$$y_1=\frac{1}{x} \;,\; y'_1=\frac{-1}{x^2} \;,\; p=\frac{3}{x}$$

$$v' = e^{-\int{}{}(\frac{-2x^{-2}}{x^{-1}}+\frac{3}{x})dx}$$

$$v' = e^{-\int{\frac{-2}{x} + \frac{3}{x}}{}dx}$$

$$v' = e^{-\int{}{}\frac{1}{x}dx}$$

$$v' = e^{-ln(x)}$$

$$v' = e^{ln(x^{-1})}$$

$$v' = e^{ln(\frac{1}{x})}$$

$$v' = \frac{1}{x}$$

Now integrate to find (v)

$$v = \int{}{}\frac{1}{x} dx$$

$$v = ln(x)$$

...and recally our definition for a second solution to our differential equation:

$$y_2 = vy_1$$

$$y_2 = ln(x)\frac{1}{x}=\frac{1}{x}ln(x)$$

...and the general solution to our differential equation is represented by:

$$y_g = C_1y_1+C_2y_2$$

$$y_G = \frac{C_1}{x} + \frac{C_2}{x}ln(x)$$