Need help solving a differential equation

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SUMMARY

The discussion focuses on solving the differential equation xy' - (x+1)y = 0 with the initial condition y(0) = 5. The participant attempts to separate variables and apply the integrating factor method, specifically using the formula dy/dx + P(x)y = Q(x). A critical error identified is the division by x, which is problematic at x=0, necessitating a careful consideration of the function's behavior around this point.

PREREQUISITES
  • Understanding of first-order linear differential equations
  • Familiarity with integrating factors and separation of variables
  • Knowledge of initial value problems
  • Basic calculus, specifically integration techniques
NEXT STEPS
  • Study the method of integrating factors for first-order linear differential equations
  • Learn about the implications of initial conditions in differential equations
  • Explore the concept of continuity and differentiability around singular points
  • Practice solving differential equations with varying initial conditions
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Students studying differential equations, educators teaching calculus, and anyone seeking to improve their problem-solving skills in mathematical analysis.

Planefreak
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Homework Statement



xy' - (x+1)y = 0, y(0) = 5

Homework Equations



Well all of the differential equation stuff seems relevant to me. But from the beginning dy/dx + P(x)y = Q(x) and yeintegral(P(x)dx) = integral(Q(x)eintegral(P(x)dx)dx + C. The separation of variables M(x)dx + N(y)dy = 0.

The Attempt at a Solution



Well I tried following separating everything out: xy' - (x+1)y = 0 y' - ((x+1)/x)y = 0/x then p(x) = -(x+1)/x and eintegral(P(x)dx) = e-integral((x+1)/x dx) and using some more math we end up with e-x/x and that is where I fall apart. Do I integrate this new function and then what? I come up with x/e-x for y and (x-1)ex for y' but when I put it back into the original equation i get x2 + x = x2 -x. Where did I go wrong?
 
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Be careful...you're dividing by x but the initial condition is for x=0. You need to consider the function on an open interval about x=0 and so you cannot divide by x.
 

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