# `2xy' - ln(x^2) = 0 , y(1) = 2` Find the particular solution that satisfies the initial condition The problem: `2xy'-ln(x^2)=0 ` is as first order ordinary differential equation that we can evaluate by applying variable separable differential equation:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Apply direct integration:` intN(y) dy= int M(x) dx` to solve for the

general solution of a differential equation.

Then, `2xy'-ln(x^2)=0` will be rearrange in to `2xy'=...

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The problem: `2xy'-ln(x^2)=0 ` is as first order ordinary differential equation that we can evaluate by applying variable separable differential equation:

`N(y)y'=M(x)`

`N(y)(dy)/(dx)=M(x)`

`N(y) dy=M(x) dx`

Apply direct integration:` intN(y) dy= int M(x) dx` to solve for the

general solution of a differential equation.

Then, `2xy'-ln(x^2)=0` will be rearrange in to `2xy'= ln(x^2)`

Let `y' = (dy)/(dx)` , we get: `2x(dy)/(dx)= ln(x^2)`

or`2x(dy)= ln(x^2)(dx)`

Divide both sides by `x` to express in a form of `N(y) dy=M(x) dx` :

`(2xdy)/x= (ln(x^2)dx)/x`

`2dy= (ln(x^2)dx)/x`

Applying direct integration, we will have:

`int 2dy= int (ln(x^2)dx)/x`

For the left side, recall `int dy = y` then `int 2dy = 2y`

For the right side, we let `u =x^2` then `du =2x dx` or `dx=(du)/(2x)` .

`int (ln(x^2))/xdx=int (ln(u))/x*(du)/(2x)`

` =int (ln(u)du)/(2x^2)`

` =int (ln(u)du)/(2u) `

` =1/2 int ln(u)/u du`

Let `v=ln(u)` then `dv = 1/udu` ,we get:

`1/2 int ln(u)/u du=1/2 int v* dv`

Applying the Power Rule of integration: `int x^n dx = x^(n+1)/(n+1)+C`

`1/2 int v* dv= 1/2 v^(1+1)/(1+1)+C`

`= 1/2*v^2/2+C`

`=1/4v^2+C`

Recall `v = ln(u)` and `u = x^2` then `v =ln(x^2)` .

The integral will be:

`int (ln(x^2))/xdx=1/4(ln(x^2))^2 +C or(ln(x^2))^2 /4+C`

Combing the results from both sides, we get the general solution of the differential equation as:

`2y = (ln(x^2))^2 /4+C`

or `y =(ln(x^2))^2 /8+C`

To solve for the arbitary constant (C), we consider the initial condition `y(1)=2`

When we plug-in the values, we get:

`2 =(ln(1^2))^2 /8+C`

`2 =0/8+C`

`2=0+C`

then `C=2`

.Plug-in `C=2` on the general solution: `y =(ln(x^2))^2 /8+C` , we get the

particular solution as:

`y =(ln(x^2))^2 /8+2`

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