Question

First change to standard form of first order linear differential equation and then by finding the appropriate integrating factor find a particular solution for the initial value problem:xy' + y = lnx ; y(e)=1

Answer 1

Answer:

Step-by-step explanation:

Given is a Differential equation as

$xy' + y = lnx ; y(e)=1$

To bring it to linear form we can divide the full equation by x

$y'+\frac{y}{x} =\frac{ln x}{x}$

This is of the form

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

p(x) = 1/x

So find

$e^{\int\limits{\frac{1}{x} } \, dx } = e^{ln x} = x$

Solution is

$xy = \int {x*lnx /x } \, dx =xln x -x +C$

Use the initial value as y(e) =1

$e= eln e -e+C\\C=e$

So solution is

$xy =xln x -x+e$