Trouble with particular solution of differential equation - rewrite my answer?

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SUMMARY

The forum discussion centers on finding the particular solution to a differential equation, specifically the equation dy/dx = e - 2x(x - 5). The initial attempt yielded the expression (1/2)ln(-x^2 + 10x), which satisfied the initial conditions but was incorrect. The correct approach involved integrating both sides, leading to the solution 2y = ln(x^2 - 10x + C), with the constant C determined to be 50. This highlights the importance of ensuring that solutions satisfy both the differential equation and initial conditions.

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cowmoo32
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Trouble with particular solution of differential equation - [SOLVED]

Homework Statement


Find the particular solution of the differential equation
c995a54ca48d21f6dc81446bdd54881.png

satisfying the initial condition
907c69d9098d2e376491d1dfd7f0771.png



The Attempt at a Solution


[STRIKE]I end up with (1/2)ln(-x^2+10x) which does satisfy the initial conditions, yet it's coming back as incorrect. Is there another way I can write the formula?[/STRIKE]

I figured it out, check post #3
 
Last edited:
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cowmoo32 said:

Homework Statement


Find the particular solution of the differential equation
c995a54ca48d21f6dc81446bdd54881.png

satisfying the initial condition
907c69d9098d2e376491d1dfd7f0771.png



The Attempt at a Solution


I end up with (1/2)ln(-x^2+10x) which does satisfy the initial conditions, yet it's coming back as incorrect. Is there another way I can write the formula?
Show how you got the result you show. I assume you mean y = (1/2)ln(-x2 + 10x).

Satisfying the initial conditions is not enough: the solution has to satisfy the differential equation, too.
 
Scratch that above, I had a negative out of place. See below. I corrected it and it's coming back as correct.

[itex]\frac{dy}{dx}[/itex]=e-2x(x-5)

[itex]\int[/itex] (e2y)dy = [itex]\int (x-5)dx[/itex]

e2y = x2 - 10x + C

2y = ln(x2 - 10x + C)

C = 50
 
Last edited:

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