Solving this ODE for an initial value problem

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SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) given by \( x \frac{du}{dx} = (u-x)^3 + u \) with the initial condition \( u(1) = 10 \). The key strategy involves a change of variables by letting \( v = u - x \), which transforms the equation into a more manageable form. The participants emphasize the need to express \( \frac{du}{dx} \) in terms of \( v \) and to substitute back to isolate \( u \) as a function of \( x \).

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with variable substitution techniques
  • Knowledge of initial value problems
  • Basic calculus concepts, including differentiation
NEXT STEPS
  • Study the method of variable separation in ODEs
  • Learn about initial value problems and their solutions
  • Explore the implications of change of variables in differential equations
  • Investigate the use of substitution methods in solving nonlinear ODEs
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Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to enhance their problem-solving skills in initial value problems.

dchau503
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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.
 
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dchau503 said:

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.
 

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