- #1

jawhnay

- 37

- 0

x

^{2}y'' - xy' = ln x

I know you have to start with x = e

^{t}or t = ln x however I'm not sure what to do next...

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- Thread starter jawhnay
- Start date

- #1

jawhnay

- 37

- 0

x

I know you have to start with x = e

- #2

Gengar

- 13

- 0

[tex]\frac{d}{dx}=e^{-t}\frac{d}{dt}, \frac{d^2}{dx^2}=e^{-2t}(\frac{d^2}{dt^2}-\frac{d}{dt})[/tex]

which when substituted into your original equation should yield an inhomogeneous 2nd order ODE with constant coefficients which we can do via the standard method of finding complementary solutions and then a particular solution.

- #3

JJacquelin

- 801

- 34

x²z'+xz=ln(x)

Solving x²z'-xz=0 leads to z=c*x where c=constant

Then bing back z=u(x)*x into x²z'-xz=ln(x)

which leads to u' =ln(x)/(x^3)

u(x)=-(2*ln(x)+1)/(4*x²)+C where C=constant

y'=z(x)=u(x)*x=-(2ln(x)+1)/(4*x)+C*x

y= -(ln(x)+1)*ln(x)/4 +A*x² +B where A, B are constants.

- #4

Mark44

Mentor

- 36,893

- 8,942

Thread closed. See the duplicate post at https://www.physicsforums.com/showthread.php?t=680980.

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