Need help with differential eq. from book

[Pollywoggy]

On page 11 of Differential Equations Demystified (Krantz), there is an example that goes from this:

<br /> <br /> e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x<br /> <br />

To this:

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

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


thanks

 

[jeffreydk]

To solve differential equations in the form

y&#039;+P(x)y=Q(x)

it is useful to use an integrating factor defined by

\mu=\exp{\left(\int P(x)dx\right)}

We multiply both sides of the equation by this,

\mu y&#039;+\mu P(x)y=\mu Q(x)

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

So then we get that

\left(\mu y\right)&#039;=\mu Q(x)

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

\left(e^{x^2} y\right)&#039;=e^{x^2}x

 

[HallsofIvy]

More simply, to see that the two expressions are the same, go from
\left[ e^{x^2} \cdot y \right]^\prime = x \cdot e^{x^2}
to
e^{x^2} \cdot y^\prime + e^{x^2} \cdot 2xy = e^{x^2} \cdot x
by doing that differentiation on the left.

 

[Pollywoggy]

To solve differential equations in the form

y&#039;+P(x)y=Q(x)

it is useful to use an integrating factor defined by

\mu=\exp{\left(\int P(x)dx\right)}

Thanks, this is what I was looking for but it was not explained in the book.