# solve the differential equation : ylnx - xy' = 0

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### 2 Answers

ylnx - xy' = 0

First we will rewrite the equations.

ylnx = xy'

==> ylnx = x *dy/dx

In solving differential equation, out first priority is to isolate x an y terms on different sides.

Then we will multiply both sides by dx and divide by xdy

==> lnx/x) dx = dy/y

Now let us integrate both sides:

==> intg (lnx/x) dx = intg dy/y

Let u= lnx

==> du = 1/x dx

==> x du = dx

==> intg u/x * xdu = ln y

==> intg u du = ln y

==> ln y = u^2 /2 + C

Now subsitute with u= ln x

==> ln y = (lnx)^2 /2 + C

**==> y= e^(1/2)(lnx)^2 + C**

To solve the differential equation ylnx - xy' = 0.

Solution:

ylnx - xy' = 0.

Rewrite this as:

ylnx = xy'

(lnx)/x = y'/y.

Or

(1/y) dy/dx =( lnx)/x

(1/y)dy = (lnx) dx/x.

Integrate both sides:

lny = (lnx)^2 /2 + C . As dln(x)/dx = 1/x.

ln y =( lnx)^2.

y = e ^ (lnx)^2 +C.

y = k(e^(lnx) ) ^lnx.

y = kx ^lnx is the solution of the gven differential equation.