Differential equation on -x^2

**Trikotnik** · September 12th 2010, 08:46 AM

The differential equation $y' + 2xy = 4e$ on $-x^2$ is given.

Solve the modified homogeneous equation.
What is the general solution to the equation?
Find that particulate solution which goes through T(0,2).

**CaptainBlack** · September 12th 2010, 11:37 PM

Originally Posted by Trikotnik

The differential equation $y' + 2xy = 4e$ on $-x^2$ is given.

Solve the modified homogeneous equation.
What is the general solution to the equation?
Find that particulate solution which goes through T(0,2).

What does "on -x^2" mean?

CB

**MathoMan** · September 13th 2010, 01:38 AM

First, I assume you ment $y'+2xy=4e^{-x^2}$ .

Anyway, the equation is a first order linear differential equation. In general it can be written down as $y' + p(x)y=q(x)$ . Using the variation of the constant method general solution formula can be found, and it goes something like this $y=\left(\int q(x)\cdot e^{-P(x)} dx+C\right)\cdot e^{P(x)}$ where $P$ is the primitive function of the function $p$ , $\qquad \int p(x) dx= P(x)+const.$ .

Using the formula should suffice to solve the problem(s).

**CaptainBlack** · September 13th 2010, 02:38 AM

Originally Posted by Trikotnik

The differential equation $y' + 2xy = 4e$ on $-x^2$ is given.

Solve the modified homogeneous equation.
What is the general solution to the equation?
Find that particulate solution which goes through T(0,2).

If as MathoMan suggests you mean:

$y'+2xy=e^{-x^2}$

we can use an integrating factor of $e^{x^2}$ to get:

$\dfrac{d}{dx}\left( e^{x^2}y(x)\right)=1$

so:

$y(x)=e^{-x^2}(x+c)$

CB

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