Solve Differential Equation dy/dx = x + y for General Solution

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SUMMARY

The general solution for the differential equation dy/dx = x + y is y = Ce^x - x - 1, where C is a constant. The verification process involves substituting the solution back into the original equation to confirm its validity. The derivative of the solution, y' = Ce^x - 1, aligns with the original equation when rearranged, demonstrating that the solution is indeed correct. This discussion clarifies the steps necessary to validate the general solution of the given differential equation.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the method of integrating factors in solving differential equations.
  • Knowledge of calculus, particularly derivatives and their applications.
  • Basic algebra skills for manipulating equations and verifying solutions.
NEXT STEPS
  • Study the method of integrating factors for solving first-order linear differential equations.
  • Learn how to verify solutions of differential equations through substitution.
  • Explore the concept of general solutions versus particular solutions in differential equations.
  • Investigate advanced topics in differential equations, such as systems of equations and non-linear equations.
USEFUL FOR

Students studying calculus, particularly those focusing on differential equations, as well as educators seeking to clarify solution verification methods for first-order linear equations.

quantum13
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I read this in my calculus first year textbook, and I am just curious.

Homework Statement


For differential equation dy/dx = x + y, the general solution is y = Ce^x - x -1 (the solution of which is beyond this course). Verify this by solving dy/dx.


Homework Equations


calculus equations


The Attempt at a Solution


dy/dx = Ce^x - 1

this looks nothing like x + y however. what am I missing?
 
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x + y = x + Ce^x-x-1=Ce^x-1

Simply substitute your solution into the original differential equation to verify.
 
Thanks! At least it's a trivial solution, which kind of makes me feel better, but not really =]
 
I will do the same on this way:

y' = x+y

y=Ce^x - x -1

y'=Ce^x-1

Ce^x-1=y+x

y=Ce^x-1-x
 

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