Need help with differential eq. from book

  • Thread starter Pollywoggy
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  • #1
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On page 11 of Differential Equations Demystified (Krantz), there is an example that goes from this:

[tex]

e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x

[/tex]

To this:

[tex]
\left[ e^{x^2} \cdot y \right]^\prime = x \cdot e^{x^2}
[/tex]

Would someone give me a hint as to how they got from the first equation to the second?


thanks
 

Answers and Replies

  • #2
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To solve differential equations in the form

[tex] y'+P(x)y=Q(x)[/tex]

it is useful to use an integrating factor defined by

[tex]\mu=\exp{\left(\int P(x)dx\right)}[/tex]

We multiply both sides of the equation by this,

[tex]\mu y'+\mu P(x)y=\mu Q(x)[/tex]

and if you look closely the left hand side is the product rule for differentiating [itex]y\mu[/itex].

So then we get that

[tex]\left(\mu y\right)'=\mu Q(x)[/tex]

And you'll see if you relate this to what you have, [itex]\mu=e^{x^2}[/itex] and then from the last equation,

[tex]\left(e^{x^2} y\right)'=e^{x^2}x[/tex]
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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More simply, to see that the two expressions are the same, go from
[tex]\left[ e^{x^2} \cdot y \right]^\prime = x \cdot e^{x^2}[/tex]
to
[tex] e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x[/tex]
by doing that differentiation on the left.
 
  • #4
53
0
To solve differential equations in the form

[tex] y'+P(x)y=Q(x)[/tex]

it is useful to use an integrating factor defined by

[tex]\mu=\exp{\left(\int P(x)dx\right)}[/tex]
Thanks, this is what I was looking for but it was not explained in the book.
 

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