The differential equation $\displaystyle y' + 2xy = 4e$ on $\displaystyle -x^2$ 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).

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- Sep 12th 2010, 08:46 AMTrikotnikDifferential equation on -x^2
The differential equation $\displaystyle y' + 2xy = 4e$ on $\displaystyle -x^2$ 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). - Sep 12th 2010, 11:37 PMCaptainBlack
- Sep 13th 2010, 01:38 AMMathoMan
First, I assume you ment $\displaystyle y'+2xy=4e^{-x^2} $.

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

Using the formula should suffice to solve the problem(s). - Sep 13th 2010, 02:38 AMCaptainBlack