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## Homework Statement

Find the solution to the differential equation. Which passes through the point (0,e).

[itex]\frac{dy}{dx}[/itex] = [itex]\frac{11xy}{(ln(y))^{10}}[/itex]

## Homework Equations

I can get through the integration part but I am lost when it comes to using the ln rules to find the solution.

## The Attempt at a Solution

First step I move the variables to each side.

[itex]\frac{(ln(y))^{10}}{y}[/itex] dy = 11x dx

I then integrate each side:

∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy = ∫11x dx

∫[itex]\frac{(ln(y))^{10}}{y}[/itex] dy, I used sub method. u = ln(y), du = 1/y dy -> dy = ydu

∫[itex]\frac{u^{10}}{y}[/itex] * y du, so the y's cancel out. making it:

∫ u[itex]^{10}[/itex] du = [itex]\frac{u^{11}}{11}[/itex] = [itex]\frac{ln(y)^{11}}{11}[/itex]

The other side is simple integration:

∫11x dx = [itex]\frac{11x^{2}}{2}[/itex], this gives:

[itex]\frac{ln(y)^{11}}{11}[/itex] = [itex]\frac{11x^{2}}{2}[/itex] + c, since it passes through

the point (0,e), I find my constant value c right away.

y(0)=e =[itex]\frac{ln(e)^{11}}{11}[/itex] = [itex]\frac{11(0)^{2}}{2}[/itex] + c

y(0)=e = 1 = 0 + c, so c = 1.

I then multiplied the 11 by both sides and got:

(ln(y))[itex]^{11}[/itex] = [itex]\frac{121x^{2}}{2}[/itex] + 11

This is where I am getting confused. I do not know how to use the ln rules on this one. The power of 11 is confusing me and I can not find an example like this one. Any help would be nice. thanks.