Need help with differential eq. from book

[Pollywoggy]

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

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

To this:

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

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

$$y'+P(x)y=Q(x)$$

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,

$$\mu y'+\mu P(x)y=\mu Q(x)$$

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

So then we get that

$$\left(\mu y\right)'=\mu Q(x)$$

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

$$\left(e^{x^2} y\right)'=e^{x^2}x$$

 

[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

$$y'+P(x)y=Q(x)$$

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.