Solving a special type of a second order differential equation

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SUMMARY

The discussion focuses on solving the second-order differential equation y''[x] = y'[x] + x using a specific method. The solution involves transforming the equation into a first-order equation by substituting v = y'[x], leading to v'[x] - v = x. The final solution is derived as y[x] = C2 - C1 * e^-x - 0.5x^2 - x. The confusion arises from the application of Type 1 equations, which typically apply when x is absent, prompting a clarification on the classification of the equation types.

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  • Understanding of second-order differential equations
  • Familiarity with first-order differential equations and transformations
  • Knowledge of integrating factors in differential equations
  • Basic concepts of exponential functions and their derivatives
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  • Study the method of solving second-order differential equations using substitution techniques
  • Learn about integrating factors and their application in first-order differential equations
  • Explore the classification of differential equations and the conditions for Type 1 and Type 2 equations
  • Practice solving various forms of second-order differential equations to reinforce understanding
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Students studying differential equations, mathematics educators, and anyone looking to deepen their understanding of solving second-order differential equations and their classifications.

Jim4592
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Homework Statement



y''[x] = y'[x] + x


Homework Equations


We were taught two special types of second order diff. equations:

Type 1: Supposed to be when x is missing
v = y'[x]
v'[x] = y''[x]

Type 2: Supposed to be when y is missing
v = y'[x]
v v'[x] = y''[x]


The Attempt at a Solution


The answer key reads:

v'[x] - v = x

d/dx{e^-x v} = x e^-x

e^-x v = C1 * e^-x -x -1

v = C1 * e^x -x -1

y[x] = c2 - c1 * e^-x - (.5)x^2 - x


I don't understand why we used type 1 to solve this problem since x is clearly stated in the problem. I was hoping someone could explain, thanks.
 
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Are you sure you have listed the two types correctly? Usually, if y is missing you would do:

v = y'
v' = y''

so your second order equation in y becomes a first order equation in v. That is the type of equation you have and also the method your manual gives.
 

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