Find the general solution?

Question

Find the general solution of xy'+(ln x)y=0.

 

Answer: y=ce^(-(ln x)^2)/2

 

Please show all your work step by step. And tell me the input on Ti-89 for this problem to find the answer. 

Answer 1

Hi Sun,

 

This is a first order linear differential equation, and to solve these equations, we must put it into the standard form

 

y' + p(x)y = q(x)

 

So, we divide out the x in the equation to yield

 

y' + ( (ln x)/x )y = 0

 

Here, p(x) = (ln x)/x, and q(x) = 0. The general solution to first order linear differential equations is given by

 

y = (∫u(x)q(x) dx + C)/u(x)

 

where u(x) is the integrating factor

 

u(x) = e^{∫p(x) dx}

 

So, to begin, we must first find the integrating factor u(x):

 

u(x) = e^{∫(ln x)/x dx}

 

       = e^{((ln x)^2)/2}  (integration by parts with ln(x) as u and 1/x dx as dv)

 

As a side note, we will work out ∫(ln x)/x dx. We use integration by parts with u₁(x) = ln x and dv = 1/x dx (I am using u₁ for the integration by parts so as to not confuse it with our integrating factor u(x)). Recall that

 

∫u₁(x)v'(x) dx = u₁(x)v(x) - ∫u'(x)v(x) dx

 

So, with u₁(x) = ln x and v'(x) = 1/x dx, we have

 

∫u1(x)v'(x) dx = ∫(ln x)/x dx

 

                      = (ln x)*(ln x) - ∫(1/x)*(ln x) dx

 

The above tells us that

 

∫(ln x)/x dx = (ln x)*(ln x) - ∫(1/x)*(ln x) dx

 

So adding the ∫(1/x)*(ln x) dx to both sides yields

 

2 ∫(1/x)*(ln x) dx = (ln x)²

 

and so

 

 ∫(1/x)*(ln x) dx = (ln x)²/2 +c

 

Back to our solution, since q(x) = 0, the numerator of our solution is just the constant term C. Our solution is therefore given as

 

y = C/e^{((ln x)^2)/2}

 

    = C e^{-((ln x)^2)/2}