Solving this ODE for an initial value problem

[dchau503]

Homework Statement

$$x \frac{du}{dx} \ = \ (u-x)^3 + u$$

solve for u(x) and use $$u(1) \ = \ 10$$ to solve for u without a constant.

Homework Equations

The given hint is to let $v=u-x$

The Attempt at a Solution

This equation is not separable and the book wants me to make it separable by a change of variables, i.e.

$$u=v+x \ \ \text{and in replacing the original equation with the hint, I get} \frac{du}{dx} = \frac{v^3+u}{x}$$.

From $$u=v+x, \ \text{I also know that} \frac{du}{dx} \ \text{is also equal to 1, so} \frac{v^3+u}{x} = 1$$

But this gets rid of all the differentials and I need guidance on how to solve for u in terms of just x.

 

[LCKurtz]

Homework Statement

$$x \frac{du}{dx} \ = \ (u-x)^3 + u$$

solve for u(x) and use $$u(1) \ = \ 10$$ to solve for u without a constant.

Homework Equations

The given hint is to let $v=u-x$

The Attempt at a Solution

This equation is not separable and the book wants me to make it separable by a change of variables, i.e.

$$u=v+x \ \ \text{and in replacing the original equation with the hint, I get} \frac{du}{dx} = \frac{v^3+u}{x}$$.

If $u=v+x$, what is $\frac {du}{dx}$ in terms of $v$? Also substitute for that $u$ that is left and show us what you get.