Question

d^2 y /dx^2 + 1/x dy/dx = 12 log x / x^2​

Answer 1

Answer:

Step-by-step explanation:

d^2y/dx^2 + 1/x dy/dx =12 logx/x^2

We have solve this differential equation,

Now let, dy/dx = z, this implies d^2y/dx^2 = dz/dx

Now putting this value in the given differential equation in place of dy/dx and d^2y/dx^2, we get,

dz/dx + 1/x z = 12logx/x^2

Now solving this differential equation,

Here the integrating factor, is, I. F. = e^∫(1/x) dx=e^logx=x

Now multiplying integrating factor the both sides of the equation we get,

Z×x=∫(12logx/x^2) ×xdx +c

This implies, zx=∫12logx/x^2dx+c

This implies, zx =12∫logx/xdx+c

So, zx = (12/2) (logx) ^2+c[ where c is constant]

Now , z=dy/dx

So xdy/dx= 6(logx) ^2+c

So, dy/dx=6(logx) ^2/x+c/x

i,e. dy = 6(logx) ^2/xdx + c/x dx

Integrating both sides we get,

Y= 2(logx) ^3+c logx+ d [where c, d are arbitrary constant]