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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

Respuesta :

Answer:

Step-by-step explanation:

Given is a Differential equation as

[tex]xy' + y = lnx ; y(e)=1[/tex]

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

[tex]y'+\frac{y}{x} =\frac{ln x}{x}[/tex]

This is of the form

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

p(x) = 1/x

So find

[tex]e^{\int\limits{\frac{1}{x} } \, dx } = e^{ln x} = x[/tex]

Solution is

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

Use the initial value as y(e) =1

[tex]e= eln e -e+C\\C=e[/tex]

So solution is

[tex]xy =xln x -x+e[/tex]

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